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15 Probability Questions And Practice Problems for Middle and High School: Harder Exam Style Questions Included

Beki Christian

Probability questions and probability problems require students to work out how likely it is that something is to happen. Probabilities can be described using words or numbers. Probabilities range from 0 to 1 and can be written as fractions, decimals or percentages .

Here you’ll find a selection of probability questions of varying difficulty showing the variety you are likely to encounter in middle school and high school, including several harder exam style questions.

What are some real life examples of probability?

The more likely something is to happen, the higher its probability. We think about probabilities all the time.

For example, you may have seen that there is a 20% chance of rain on a certain day or thought about how likely you are to roll a 6 when playing a game, or to win in a raffle when you buy a ticket.

Probability Questions For Middle & High School

Get all the questions from this blog in an easy-to-download format, including answer key. Includes a mix of worded problems and deeper problem solving questions.

How to calculate probabilities

The probability of something happening is given by:

We can also use the following formula to help us calculate probabilities and solve problems:

• Probability of something not occuring = 1 – probability of if occurring P(not\;A) = 1 - P(A)
• For mutually exclusive events: Probability of event A OR event B occurring = Probability of event A + Probability of event B P(A\;or\;B) = P(A)+P(B)
• For independent events: Probability of event A AND event B occurring = Probability of event A times probability of event B P(A\;and\;B) = P(A) × P(B)

Probability question: A worked example

Question: What is the probability of getting heads three times in a row when flipping a coin?

When flipping a coin, there are two possible outcomes – heads or tails. Each of these options has the same probability of occurring during each flip. The probability of either heads or tails on a single coin flip is ½.

Since there are only two possible outcomes and they have the same probability of occurring, this is called a binomial distribution.

Let’s look at the possible outcomes if we flipped a coin three times.

Let H=heads and T=tails.

The possible outcomes are: HHH, THH, THT, HTT, HHT, HTH, TTH, TTT

Each of these outcomes has a probability of ⅛.

Therefore, the probability of flipping a coin three times in a row and having it land on heads all three times is ⅛.

Middle school probability questions

In middle school, probability questions introduce the idea of the probability scale and the fact that probabilities sum to one. We look at theoretical and experimental probability as well as learning about sample space diagrams and venn diagrams.

6th grade probability questions

1. Which number could be added to this spinner to make it more likely that the spinner will land on an odd number than a prime number?

Currently there are two odd numbers and two prime numbers so the chances of landing on an odd number or a prime number are the same. By adding 3, 5 or 11 you would be adding one prime number and one odd number so the chances would remain equal.

By adding 9 you would be adding an odd number but not a prime number. There would be three odd numbers and two prime numbers so the spinner would be more likely to land on an odd number than a prime number.

2. Ifan rolls a fair dice, with sides labeled A, B, C, D, E and F. What is the probability that the dice lands on a vowel?

A and E are vowels so there are 2 outcomes that are vowels out of 6 outcomes altogether.

Therefore the probability is   \frac{2}{6} which can be simplified to \frac{1}{3} .

7th grade probability questions

3. Max tested a coin to see whether it was fair. The table shows the results of his coin toss experiment:

Heads          Tails

    26                  41

What is the relative frequency of the coin landing on heads?

Max tossed the coin 67 times and it landed on heads 26 times.

\text{Relative frequency (experimental probability) } = \frac{\text{number of successful trials}}{\text{total number of trials}} = \frac{26}{67}

4. Grace rolled two dice. She then did something with the two numbers shown. Here is a sample space diagram showing all the possible outcomes:

What did Grace do with the two numbers shown on the dice?

Add them together

Subtract the number on dice 2 from the number on dice 1

Multiply them

Subtract the smaller number from the bigger number

For each pair of numbers, Grace subtracted the smaller number from the bigger number.

For example, if she rolled a 2 and a 5, she did 5 − 2 = 3.

8th grade probability questions

5. Alice has some red balls and some black balls in a bag. Altogether she has 25 balls. Alice picks one ball from the bag. The probability that Alice picks a red ball is x and the probability that Alice picks a black ball is 4x. Work out how many black balls are in the bag.

Since the probability of mutually exclusive events add to 1:  

\begin{aligned} x+4x&=1\\\\ 5x&=1\\\\ x&=\frac{1}{5} \end{aligned}

\frac{1}{5} of the balls are red and \frac{4}{5} of the balls are blue.

6. Arthur asked the students in his class whether they like math and whether they like science. He recorded his results in the venn diagram below.

How many students don’t like science?

We need to look at the numbers that are not in the ‘Like science’ circle. In this case it is 9 + 7 = 16.

High school probability questions

In high school, probability questions involve more problem solving to make predictions about the probability of an event. We also learn about probability tree diagrams, which can be used to represent multiple events, and conditional probability.

9th grade probability questions

7. A restaurant offers the following options:

Starter – soup or salad

Main – chicken, fish or vegetarian

Dessert – ice cream or cake

How many possible different combinations of starter, main and dessert are there?

The number of different combinations is 2 × 3 × 2 = 12.

8. There are 18 girls and 12 boys in a class. \frac{2}{9} of the girls and \frac{1}{4} of the boys walk to school. One of the students who walks to school is chosen at random. Find the probability that the student is a boy. 

First we need to work out how many students walk to school:

\frac{2}{9} \text{ of } 18 = 4

\frac{1}{4} \text{ of } 12 = 3

7 students walk to school. 4 are girls and 3 are boys. So the probability the student is a boy is \frac{3}{7} .

9. Rachel flips a biased coin. The probability that she gets two heads is 0.16. What is the probability that she gets two tails?

We have been given the probability of getting two heads. We need to calculate the probability of getting a head on each flip.

Let’s call the probability of getting a head p.

The probability p, of getting a head AND getting another head is 0.16.

Therefore to find p:

The probability of getting a head is 0.4 so the probability of getting a tail is 0.6.

The probability of getting two tails is 0.6 × 0.6 = 0.36 .

10th grade probability questions

10. I have a big tub of jelly beans. The probability of picking each different color of jelly bean is shown below:

If I were to pick 60 jelly beans from the tub, how many orange jelly beans would I expect to pick?

First we need to calculate the probability of picking an orange. Probabilities sum to 1 so 1 − (0.2 + 0.15 + 0.1 + 0.3) = 0.25.

The probability of picking an orange is 0.25.

The number of times I would expect to pick an orange jelly bean is 0.25 × 60 = 15 .

11. Dexter runs a game at a fair. To play the game, you must roll a dice and pick a card from a deck of cards.

To win the game you must roll an odd number and pick a picture card. The game can be represented by the tree diagram below.

Dexter charges players $1 to play and gives $3 to any winners. If 260 people play the game, how much profit would Dexter expect to make?

Completing the tree diagram:

Probability of winning is \frac{1}{2} \times \frac{4}{13} = \frac{4}{26}

If 260 play the game, Dexter would receive $260.

The expected number of winners would be \frac{4}{26} \times 260 = 40

Dexter would need to give away 40 × $3 = $120 .

Therefore Dexter’s profit would be $260 − $120 = $140.

12. A fair coin is tossed three times. Work out the probability of getting two heads and one tail.

There are three ways of getting two heads and one tail: HHT, HTH or THH.

The probability of each is \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} 

Therefore the total probability is \frac{1}{8} +\frac{1}{8} + \frac{1}{8} = \frac{3}{8}

11th/12th grade probability questions

13. 200 people were asked about which athletic event they thought was the most exciting to watch. The results are shown in the table below.

A person is chosen at random. Given that that person chose 100m, what is the probability that the person was female?

Since we know that the person chose 100m, we need to include the people in that column only.

In total 88 people chose 100m so the probability the person was female is \frac{32}{88}   .

14.   Sam asked 50 people whether they like vegetable pizza or pepperoni pizza.

37 people like vegetable pizza. 

25 people like both. 

3 people like neither.

Sam picked one of the 50 people at random. Given that the person he chose likes pepperoni pizza, find the probability that they don’t like vegetable pizza.

We need to draw a venn diagram to work this out.

We start by putting the 25 who like both in the middle section. The 37 people who like vegetable pizza includes the 25 who like both, so 12 more people must like vegetable pizza. 3 don’t like either. We have 50 – 12 – 25 – 3 = 10 people left so this is the number that must like only pepperoni.

There are 35 people altogether who like pepperoni pizza. Of these, 10 do not like vegetable pizza. The probability is   \frac{10}{35} .

15. There are 12 marbles in a bag. There are n red marbles and the rest are blue marbles. Nico takes 2 marbles from the bag. Write an expression involving n for the probability that Nico takes one red marble and one blue marble.

We need to think about this using a tree diagram. If there are 12 marbles altogether and n are red then 12-n are blue.

To get one red and one blue, Nico could choose red then blue or blue then red so the probability is:

Looking for more middle school and high school probability math questions?

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The content in this article was originally written by secondary teacher Beki Christian and has since been revised and adapted for US schools by elementary math teacher Katie Keeton.

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Probability - Problem Solving

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• Geoff Pilling
• Sandeep Bhardwaj

To solve problems on this page, you should be familiar with

• Uniform Probability
• Probability - By Outcomes
• Probability - Rule of Sum
• Probability - Rule of Product
• Probability - By Complement
• Probability - Independent Events
• Conditional Probability

Problem Solving - Basic

Problem solving - intermediate, problem solving - difficult.

If I throw 2 standard 5-sided dice, what is the probability that the sum of their top faces equals to 10? Assume both throws are independent to each other. Solution : The only way to obtain a sum of 10 from two 5-sided dice is that both die shows 5 face up. Therefore, the probability is simply \( \frac15 \times \frac15 = \frac1{25} = .04\)

If from each of the three boxes containing \(3\) white and \(1\) black, \(2\) white and \(2\) black, \(1\) white and \(3\) black balls, one ball is drawn at random. Then the probability that \(2\) white and \(1\) black balls will be drawn is?

2 fair 6-sided dice are rolled. What is the probability that the sum of these dice is \(10\)? Solution : The event for which I obtain a sum of 10 is \(\{(4,6),(6,4),(5,5) \}\). And there is a total of \(6^2 = 36\) possible outcomes. Thus the probability is simply \( \frac3{36} = \frac1{12} \approx 0.0833\)

If a fair 6-sided dice is rolled 3 times, what is the probability that we will get at least 1 even number and at least 1 odd number?

Three fair cubical dice are thrown. If the probability that the product of the scores on the three dice is \(90\) is \(\dfrac{a}{b}\), where \(a,b\) are positive coprime integers, then find the value of \((b-a)\).

You can try my other Probability problems by clicking here

Suppose a jar contains 15 red marbles, 20 blue marbles, 5 green marbles, and 16 yellow marbles. If you randomly select one marble from the jar, what is the probability that you will have a red or green marble? First, we can solve this by thinking in terms of outcomes. You could draw a red, blue, green, or yellow marble. The probability that you will draw a green or a red marble is \(\frac{5 + 15}{5+15+16+20}\). We can also solve this problem by thinking in terms of probability by complement. We know that the marble we draw must be blue, red, green, or yellow. In other words, there is a probability of 1 that we will draw a blue, red, green, or yellow marble. We want to know the probability that we will draw a green or red marble. The probability that the marble is blue or yellow is \(\frac{16 + 20}{5+15+16+20}\). , Using the following formula \(P(\text{red or green}) = 1 - P(\text{blue or yellow})\), we can determine that \(P(\text{red or green}) = 1 - \frac{16 + 20}{5+15+16+20} = \frac{5 + 15}{5+15+16+20}\).

Two players, Nihar and I, are playing a game in which we alternate tossing a fair coin and the first player to get a head wins. Given that I toss first, the probability that Nihar wins the game is \(\dfrac{\alpha}{\beta}\), where \(\alpha\) and \(\beta\) are coprime positive integers.

Find \(\alpha + \beta\).

If I throw 3 fair 5-sided dice, what is the probability that the sum of their top faces equals 10? Solution : We want to find the total integer solution for which \(a +b+c=10 \) with integers \(1\leq a,b,c \leq5 \). Without loss of generality, let \(a\leq b \leq c\). We list out the integer solutions: \[ (1,4,5),(2,3,5), (2,4,4), (3,3,4) \] When relaxing the constraint of \(a\leq b \leq c\), we have a total of \(3! + 3! + \frac{3!}{2!} + \frac{3!}{2!} = 18 \) solutions. Because there's a total of \(5^3 = 125\) possible combinations, the probability is \( \frac{18}{125} = 14.4\%. \ \square\)

Suppose you and 5 of your friends each brought a hat to a party. The hats are then put into a large box for a random-hat-draw. What is the probability that nobody selects his or her own hat?

How many ways are there to choose exactly two pets from a store with 8 dogs and 12 cats? Since we haven't specified what kind of pets we pick, we can choose any animal for our first pick, which gives us \( 8+12=20\) options. For our second choice, we have 19 animals left to choose from. Thus, by the rule of product, there are \( 20 \times 19 = 380 \) possible ways to choose exactly two pets. However, we have counted every pet combination twice. For example, (A,B) and (B,A) are counted as two different choices even when we have selected the same two pets. Therefore, the correct number of possible ways are \( {380 \over 2} = 190 \)

A bag contains blue and green marbles. If 5 green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be 75/83 . If instead 7 blue marbles are added to the bag, the probability of drawing a blue marble would be 3/19 . What was the number of blue marbles in the bag before any changes were made?

Bob wants to keep a good-streak on Brilliant, so he logs in each day to Brilliant in the month of June. But he doesn't have much time, so he selects the first problem he sees, answers it randomly and logs out, despite whether it is correct or incorrect.

Assume that Bob answers all problems with \(\frac{7}{13}\) probability of being correct. He gets only 10 problems correct, surprisingly in a row, out of the 30 he solves. If the probability that happens is \(\frac{p}{q}\), where \(p\) and \(q\) are coprime positive integers, find the last \(3\) digits of \(p+q\).

Out of 10001 tickets numbered consecutively, 3 are drawn at random .

Find the chance that the numbers on them are in Arithmetic Progression .

The answer is of the form \( \frac{l}{k} \) .

Find \( k - l \) where \(k\) and \(l\) are co-prime integers.

HINT : You might consider solving for \(2n + 1\) tickets .

You can try more of my Questions here .

A bag contains a blue ball, some red balls, and some green balls. You reach into​ the bag and pull out three balls at random. The probability you pull out one of each color is exactly 3%. How many balls were initially in the bag?

More probability questions

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Amanda decides to practice shooting hoops from the free throw line. She decides to take 100 shots before dinner.

Her first shot has a 50% chance of going in.

But for Amanda, every time she makes a shot, it builds her confidence, so the probability of making the next shot goes up, But every time she misses, she gets discouraged so the probability of her making her next shot goes down.

In fact, after \(n\) shots, the probability of her making her next shot is given by \(P = \dfrac{b+1}{n+2}\), where \(b\) is the number of shots she has made so far (as opposed to ones she has missed).

So, after she has completed 100 shots, if the probability she has made exactly 83 of them is \(\dfrac ab\), where \(a\) and \(b\) are coprime positive integers, what is \(a+b\)?

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Probabilities: Problems with Solutions

Solved Probability Problems

Solved probability problems and solutions are given here for a concept with clear understanding.

Students can get a fair idea on the probability questions which are provided with the detailed step-by-step answers to every question.

Solved probability problems with solutions :

The graphic above shows a container with 4 blue triangles, 5 green squares and 7 red circles. A single object is drawn at random from the container.

Match the following events with the corresponding probabilities:

Number of blue triangles in a container = 4

Number of green squares = 5

Number of red circles = 7

Total number of objects = 4 + 5 + 7 = 16

(i) The objects is not a circle:

P(the object is a circle)

= Number of circles/Total number of objects

P(the object is not a circle)

= 1 - P(the object is a circle)

= (16 - 7)/16

(ii) The objects is a triangle:

P(the object is a triangle)

= Number of triangle/Total number of objects

(iii) The objects is not a triangle:

= Number of triangles/Total number of objects

P(the object is not a triangle)

= 1 - P(the object is a triangle)

= (16 - 4)/16

(iv) The objects is not a square:

P(the object is a square)

= Number of squares/Total number of objects

P(the object is not a square)

= 1 - P(the object is a square)

= (16 - 5)/16

(v) The objects is a circle:

(vi) The objects is a square:

Match the following events with the corresponding probabilities are shown below:

2. A single card is drawn at random from a standard deck of 52 playing cards.

Match each event with its probability.

Note: fractional probabilities have been reduced to lowest terms. Consider the ace as the highest card.

Total number of playing cards = 52

(i) The card is a diamond:

Number of diamonds in a deck of 52 cards = 13

P(the card is a diamond)

= Number of diamonds/Total number of playing cards

(ii) The card is a red king:

Number of red king in a deck of 52 cards = 2

P(the card is a red king)

= Number of red kings/Total number of playing cards

(iii) The card is a king or queen:

Number of kings in a deck of 52 cards = 4

Number of queens in a deck of 52 cards = 4

Total number of king or queen in a deck of 52 cards = 4 + 4 = 8

P(the card is a king or queen)

= Number of king or queen/Total number of playing cards

(iv) The card is either a red card or an ace:

Total number of red card or an ace in a deck of 52 cards = 28

P(the card is either a red card or an ace)

= Number of cards which is either a red card or an ace/Total number of playing cards

(v) The card is not a king:

P(the card is a king)

= Number of kings/Total number of playing cards

P(the card is not a king)

= 1 - P(the card is a king)

= (13 - 1)/13

(vi) The card is a five or lower:

Number of cards is a five or lower = 16

P(the card is a five or lower)

= Number of card is a five or lower/Total number of playing cards

(vii) The card is a king:

(viii) The card is black:

Number of black cards in a deck of 52 cards = 26

P(the card is black)

= Number of black cards/Total number of playing cards

3. A bag contains 3 red balls and 4 black balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is 

(ii) not black.

(i) Total number of possible outcomes = 3 + 4 = 7.

Number of favourable outcomes for the event E.

                              = Number of black balls = 4.

So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Total Number of Possible Outcomes}}\)

             = \(\frac{4}{7}\).

(ii) The event of the ball being not black = \(\bar{E}\).

Hence, required probability = P(\(\bar{E}\))

                                        = 1 - P(E)

                                        = 1 - \(\frac{4}{7}\)

                                        = \(\frac{3}{7}\).

4. If the probability of Serena Williams a particular tennis match is 0.86, what is the probability of her losing the match?

Let E = the event of Serena Williams winning.

From the question, P(E) = 0.86.

Clearly, \(\bar{E}\) = the event of Serena Williams losing.

So, P(\(\bar{E}\)) = 1 - P(E) 

                            = 1 - 0.86

                            = 0.14

                            = \(\frac{14}{100}\)

                            = \(\frac{7}{50}\).

5. Find the probability of getting 53 Sunday in a leap year.

A leap year has 366 days. So, it has 52 weeks and 2 days.

So, 52 Sundays are assured. For 53 Sundays, one of the two remaining days must be a Sunday. 

For the remaining 2 days we can have

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).

So, total number of possible outcomes = 7.

Number of favourable outcomes for the event E = 2,   [namely, (Sunday, Monday), (Saturday, Sunday)].

So, by definition: P(E) = \(\frac{2}{7}\).

6. A lot of 24 bulbs contains 25% defective bulbs. A bulb is drawn at random from the lot. It is found to be not defective and it is not put back. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

25% of 24 = \(\frac{25}{100}\) × 24 = 6.

So, there are 6 defective bulbs and 18 bulbs are not defective. 

After the first draw, the lot is left with 6 defective bulbs and 17 non-defective bulbs.

So, when the second bulb is drwn, the total number of possible outcomes = 23  (= 6+ 17).

Number of favourable outcomes for the event E = number of non-defective bulbs = 17.

So, the required probability = P(E) = (\frac{17}{23}\).

The examples can help the students to practice more questions on probability by following the concept provided in the solved probability problems.

• Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Probability for Rolling Three Dice

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• Online Practice Tests

Probability Practice Problems

1. on a six-sided die, each side has a number between 1 and 6. what is the probability of throwing a 3 or a 4, 2. three coins are tossed up in the air, one at a time. what is the probability that two of them will land heads up and one will land tails up, 3. a two-digit number is chosen at random. what is the probability that the chosen number is a multiple of 7, 4. a bag contains 14 blue, 6 red, 12 green, and 8 purple buttons. 25 buttons are removed from the bag randomly. how many of the removed buttons were red if the chance of drawing a red button from the bag is now 1/3, 5. there are 6 blue marbles, 3 red marbles, and 5 yellow marbles in a bag. what is the probability of selecting a blue or red marble on the first draw, 6. using a six-sided die, carlin has rolled a six on each of 4 successive tosses. what is the probability of carlin rolling a six on the next toss, 7. a regular deck of cards has 52 cards. assuming that you do not replace the card you had drawn before the next draw, what is the probability of drawing three aces in a row.

• 1 in 132600

8. An MP3 player is set to play songs at random from the fifteen songs it contains in memory. Any song can be played at any time, even if it is repeated. There are 5 songs by Band A, 3 songs by Band B, 2 by Band C, and 5 by Band D. If the player has just played two songs in a row by Band D, what is the probability that the next song will also be by Band D?

• Not enough data to determine.

9. Referring again to the MP3 player described in Question 8, what is the probability that the next two songs will both be by Band B?

10. if a bag of balloons consists of 47 white balloons, 5 yellow balloons, and 10 black balloons, what is the approximate likelihood that a balloon chosen randomly from the bag will be black, 11. in a lottery game, there are 2 winners for every 100 tickets sold on average. if a man buys 10 tickets, what is the probability that he is a winner, answers and explanations.

1.  B:  On a six-sided die, the probability of throwing any number is 1 in 6. The probability of throwing a 3 or a 4 is double that, or 2 in 6. This can be simplified by dividing both 2 and 6 by 2.

Therefore, the probability of throwing either a 3 or 4 is 1 in 3.

2.  D:  Shown below is the sample space of possible outcomes for tossing three coins, one at a time. Since there is a possibility of two outcomes (heads or tails) for each coin, there is a total of 2*2*2=8 possible outcomes for the three coins altogether. Note that H represents heads and T represents tails:

HHH HHT HTT HTH TTT TTH THT THH

Notice that out of the 8 possible outcomes, only 3 of them (HHT, HTH, and THH) meet the desired condition that two coins land heads up and one coin lands tails up. Probability, by definition, is the number of desired outcomes divided by the number of possible outcomes. Therefore, the probability of two heads and one tail is 3/8, Choice D.

3.  E:  There are 90 two-digit numbers (all integers from 10 to 99). Of those, there are 13 multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.

4.  B:  Add the 14 blue, 6 red, 12 green, and 8 purple buttons to get a total of 40 buttons. If 25 buttons are removed, there are 15 buttons remaining in the bag. If the chance of drawing a red button is now 1/3, then 5 of the 15 buttons remaining must be red. The original total of red buttons was 6. So, one red button was removed.

5.  D:  Use this ratio for probability:

Probability = Number of Desired Outcomes

Number of Possible Outcomes

There are 6 blue marbles and 3 red marbles for a total of 9 desired outcomes. Add the total number of marbles to get the total number of possible outcomes, 14. The probability that a red or blue marble will be selected is 9/14.

6.  C:  The outcomes of previous rolls do not affect the outcomes of future rolls. There is one desired outcome and six possible outcomes. The probability of rolling a six on the fifth roll is 1/6, the same as the probability of rolling a six on any given individual roll.

7.  D:  The probability of getting three aces in a row is the product of the probabilities for each draw. For the first ace, that is 4 in 52 or 1 in 13; for the second, it is 3 in 51 or 1 in 27; and for the third, it is 2 in 50 or 1 in 25. So the overall probability,  P , is P=1/13*1/17*1/25=1/5,525

8.  B:  The probability of playing a song by a particular band is proportional to the number of songs by that band divided by the total number of songs, or 5/15=1/3 for B and D. The probability of playing any particular song is not affected by what has been played previously, since the choice is random and songs may be repeated.

9.  A:  Since 3 of the 15 songs are by Band B, the probability that any one song will be by that band is 3/15=1/5. The probability that the next two songs are by Band B is equal to the product of two probabilities, where each probability is that the next song is by Band B: 1/5*1/5=1/25 The same probability of 1/5 may be multiplied twice because whether or not the first song is by Band B has no impact on whether the second song is by Band B. They are independent events.

10.  B:  First, calculate the total number of balloons in the bag: 47 + 5 + 10 = 62.

Ten of these are black, so divide this number by 62. Then, multiply by 100 to express the probability as a percentage:

10 / 62 = 0.16

0.16 100 = 16%

11. C: First, simplify the winning rate. If there are 2 winners for every 100 tickets, there is 1 winner for every 50 tickets sold. This can be expressed as a probability of 1/50 or 0.02. In order to account for the (unlikely) scenarios of more than a single winning ticket, calculate the probability that none of the tickets win and then subtract that from 1. There is a probability of 49/50 that a given ticket will not win. For all ten to lose that would be (49/50)^(10) ≈ 0.817. Therefore, the probability that at least one ticket wins is 1 − 0.817 = 0.183 or about 18.3%

Probability Word Problems

In these lessons, we will learn how to solve a variety of probability problems.

Related Pages Probability Tree Diagrams Probability Without Replacement Theoretical vs. Experimental Probability More Lessons On Probability

Here we shall be looking into solving probability word problems involving:

• Probability and Sample Space
• Probability and Frequency Table
• Probability and Area
• Probability of Simple Events
• Probability and Permutations
• Probability and Combinations
• Probability of Independent Events

We will now look at some examples of probability problems.

Example: At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are lorries. If every vehicle is equally likely to leave, find the probability of: a) a van leaving first. b) a lorry leaving first. c) a car leaving second if either a lorry or van had left first.

Solution: a) Let S be the sample space and A be the event of a van leaving first. n(S) = 100 n(A) = 30

c) If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars. Let T be the sample space and C be the event of a car leaving. n(T) = 99 n(C) = 60

Example: A survey was taken on 30 classes at a school to find the total number of left-handed students in each class. The table below shows the results:

A class was selected at random. a) Find the probability that the class has 2 left-handed students. b) What is the probability that the class has at least 3 left-handed students? c) Given that the total number of students in the 30 classes is 960, find the probability that a student randomly chosen from these 30 classes is left-handed.

a) Let S be the sample space and A be the event of a class having 2 left-handed students. n(S) = 30 n(A) = 5

b) Let B be the event of a class having at least 3 left-handed students. n(B) = 12 + 8 + 2 = 22

c) First find the total number of left-handed students:

Total no. of left-handed students = 2 + 10 + 36 + 32 + 10 = 90

Here, the sample space is the total number of students in the 30 classes, which was given as 960.

Let T be the sample space and C be the event that a student is left-handed. n(T) = 960 n(C) = 90

Probability And Area

Example: ABCD is a square. M is the midpoint of BC and N is the midpoint of CD. A point is selected at random in the square. Calculate the probability that it lies in the triangle MCN.

Area of square = 2x × 2x = 4x 2

This video shows some examples of probability based on area.

Probability Of Simple Events

The following video shows some examples of probability problems. A few examples of calculating the probability of simple events.

• What is the probability of the next person you meeting having a phone number that ends in 5?
• What is the probability of getting all heads if you flip 3 coins?
• What is the probability that the person you meet next has a birthday in February? (Non-leap year)

This video introduces probability and gives many examples to determine the probability of basic events.

A bag contains 8 marbles numbered 1 to 8 a. What is the probability of selecting a 2 from the bag? b. What is the probability of selecting an odd number? c. What is the probability of selecting a number greater than 6?

Using a standard deck of cards, determine each probability. a. P(face card) b. P(5) c. P(non face card)

Using Permutations To Solve Probability Problems

This video shows how to evaluate factorials, how to use permutations to solve probability problems, and how to determine the number of permutations with indistinguishable items.

A permutation is an arrangement or ordering. For a permutation, the order matters.

• If a class has 28 students, how many different arrangements can 5 students give a presentation to the class?
• How many ways can the letters of the word PHEONIX be arranged?
• How many ways can you order 3 blue marbles, 4 red marbles and 5 green marbles? Marbles of the same color look identical.

Using Combinations To Solve Probability Problems

This video shows how to evaluate combinations and how to use combinations to solve probability problems.

A combination is a grouping or subset of items. For a combination, the order does not matter.

• The soccer team has 20 players. There are always 11 players on the field. How many different groups of players can be in the field at the same time?
• A student needs 8 more classes to complete her degree. If she has met the prerequisites for all the courses, how many ways can she take 4 class next semester?
• There are 4 men and 5 women in a small office. The customer wants a site visit from a group of 2 men and 2 women. How many different groups van be formed from the office?

How To Find The Probability Of Different Events?

This video explains how to determine the probability of different events. This can be found that can be found using combinations and basic probability.

• The probability of drawing 2 cards that are both face cards.
• The probability of drawing 2 cards that are both aces.
• The probability of drawing 4 cards all from the same suite.

A group of 10 students made up of 6 females and 4 males form a committee of 4. What is the probability the committee is all male? What is the probability that the committee is all female? What is the probability the committee is made up of 2 females and 2 males?

How To Find The Probability Of Multiple Independent Events?

This video explains the counting principle and how to determine the number of ways multiple independent events can occur.

• How many ways can students answer a 3-question true of false quiz?
• How many passwords using 6 digits where the first digit must be letters and the last four digits must be numbers?
• A restaurant offers a dinner special in which you get to pick 1 item from 4 different categories. How many different meals are possible?
• A door lock on a classroom requires entry of 4 digits. All digits must be numbers, but the digits can not be repeated. How many unique codes are possible?

How To Find The Probability Of A Union Of Two Events?

This video shows how to determine the probability of a union of two events.

• If you roll 2 dice at the same time, what is the probability the sum is 6 or a pair of odd numbers?
• What is the probability of selecting 1 card that is red or a face card?

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Probability

How likely something is to happen.

Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.

Tossing a Coin

When a coin is tossed, there are two possible outcomes:

Heads (H) or Tails (T)

• the probability of the coin landing H is ½
• the probability of the coin landing T is ½

Throwing Dice

When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6 .

The probability of any one of them is 1 6

In general:

Probability of an event happening = Number of ways it can happen Total number of outcomes

Example: the chances of rolling a "4" with a die

Number of ways it can happen: 1 (there is only 1 face with a "4" on it)

Total number of outcomes: 6 (there are 6 faces altogether)

So the probability = 1 6

Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?

Number of ways it can happen: 4 (there are 4 blues)

Total number of outcomes: 5 (there are 5 marbles in total)

So the probability = 4 5 = 0.8

Probability Line

We can show probability on a Probability Line :

Probability is always between 0 and 1

Probability is Just a Guide

Probability does not tell us exactly what will happen, it is just a guide

Example: toss a coin 100 times, how many Heads will come up?

Probability says that heads have a ½ chance, so we can expect 50 Heads .

But when we actually try it we might get 48 heads, or 55 heads ... or anything really, but in most cases it will be a number near 50.

Learn more at Probability Index .

Some words have special meaning in Probability:

Experiment : a repeatable procedure with a set of possible results.

Example: Throwing dice

We can throw the dice again and again, so it is repeatable.

The set of possible results from any single throw is {1, 2, 3, 4, 5, 6}

Outcome: A possible result.

Example: "6" is one of the outcomes of a throw of a die.

Trial: A single performance of an experiment.

Example: I conducted a coin toss experiment. After 4 trials I got these results:

Three trials had the outcome "Head", and one trial had the outcome "Tail"

Sample Space: all the possible outcomes of an experiment.

Example: choosing a card from a deck

There are 52 cards in a deck (not including Jokers)

So the Sample Space is all 52 possible cards : {Ace of Hearts, 2 of Hearts, etc... }

The Sample Space is made up of Sample Points:

Sample Point: just one of the possible outcomes

Example: Deck of Cards

• the 5 of Clubs is a sample point
• the King of Hearts is a sample point

"King" is not a sample point. There are 4 Kings, so that is 4 different sample points.

There are 6 different sample points in that sample space.

Event: one or more outcomes of an experiment

Example Events:

An event can be just one outcome:

• Getting a Tail when tossing a coin
• Rolling a "5"

An event can include more than one outcome:

• Choosing a "King" from a deck of cards (any of the 4 Kings)
• Rolling an "even number" (2, 4 or 6)

Hey, let's use those words, so you get used to them:

Example: Alex wants to see how many times a "double" comes up when throwing 2 dice.

The Sample Space is all possible Outcomes (36 Sample Points):

{1,1} {1,2} {1,3} {1,4} ... ... ... {6,3} {6,4} {6,5} {6,6}

The Event Alex is looking for is a "double", where both dice have the same number. It is made up of these 6 Sample Points :

{1,1} {2,2} {3,3} {4,4} {5,5} and {6,6}

These are Alex's Results:

 After 100 Trials , Alex has 19 "double" Events ... is that close to what you would expect?

Probability

Probability is traditionally considered one of the most difficult areas of mathematics , since probabilistic arguments often come up with apparently paradoxical or counterintuitive results. Examples include the Monty Hall paradox and the birthday problem . Probability can be loosely defined as the chance that an event will happen.

• 1 Video For Beginners!
• 2 Introductory Probability
• 3.1 Types of Probability
• 3.2.1 Introductory
• 3.2.2 Intermediate
• 4 Resources

Video For Beginners!

https://youtu.be/OOdK-nOzaII?t=979

Introductory Probability

Before reading about the following topics, a student learning about probability should learn about introductory counting techniques.

• dependent probability
• independent probability

Formal Definition of Probability

The foundations of probability reside in an area of analysis known as measure theory . Measure theory in general deals with integration , in particular, how to define and extend the notion of "area" or "volume." Intuitively, therefore, probability could be said to consider how much "volume" an event takes up in a space of outcomes. Measure theory does assume considerable mathematical maturity, so it is usually ignored until one reaches an advanced undergraduate level. Once measure theory is covered, probability does become a lot easier to use and understand.

We can interpret this as saying that the event of getting Heads, and the event of getting Tails, each take up an equal half of the set of possible outcomes; the event of getting Heads or Tails is certain, and likewise the event of getting neither Heads nor Tails has probability 0.

Of course, to understand this example doesn't need measure theory, but it does show how to translate a very basic situation into measure-theoretic language. Furthermore, if one wanted to determine whether the coin was fair or weighted, it would be difficult to do that without using inferential methods derived from measure theory.

Types of Probability

Part of a comprehensive understanding of basic probability includes an understanding of the differences between different kinds of probability problems.

• algebraic probability
• combinatorial probability problems involve counting outcomes.
• geometric probability

Important subdivisions of probability include

• stochastic processes
• mathematical statistics

Example Problems

Introductory.

• 2006 AMC 10B Problem 17
• 2006 AMC 10B Problem 21

Intermediate

• 2006 AIME II Problem 5
• 2007 AIME II Problem 10
• Introduction to Counting and Probability by David Patrick
• Intermediate Counting and Probability by David Patrick
• Combinatorics
• Mathematics

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Probabilistic World

Probability Questions from the Real World (With Simulations)

Posted on August 2, 2020 Written by The Cthaeh Leave a Comment

Welcome to my introductory post to a large series that I’m starting today. The main purpose of this post is to get you in the mood for the posts to follow. Namely, exploring and solving interesting probability questions from the real world.

Most of my posts so far have been more on the theoretical side. In previous posts, I introduced important concepts from probability theory (and related fields like statistics and combinatorics):

• probabilities and sample spaces
• the law of large numbers and expected value
• permutations, combinations, and other combinatorics concepts
• mean and variance
• probability distributions
• Bayes’ theorem

And I’m going to continue introducing more concepts in the future, basic and advanced concepts alike. I personally find them fascinating in their own right. In the context of mathematics, they are interesting, thought provoking, and (some would say) even beautiful.

But they’re not just interesting, they’re also extremely useful. And when I say useful, I don’t just mean useful for mathematicians and scientists. I would argue they are a potentially useful tool for everybody.

Table of Contents

Skills in the real world

Think about the following skills for a moment:

• digesting nutrition
• absorbing water

You might find it strange that I’m calling these “skills”, but essentially they are. Of course, they are skills related to basic biological survival and, by definition, (almost) every living organism needs to have them in order to remain such.

Now, what about skills like:

• being fluent in a popular language
• detecting misinformation
• imagination and creativity
• performing CPR

In my opinion, skills like these (and many more) are always good to have. Regardless of your job, your age, or where you live. They are obviously not as essential as the previous category, but all of them are things that will allow you to achieve outcomes and take advantage of situations which you otherwise might not be able to.

Are probability theory skills useful in the real world?

So, where does having probability theory skills fit in all this? Well, I think it easily fits in the second category, though this isn’t as obviously true as some of the other skills in the list above. But think about it, what is probability theory really about? What does the ability to accurately calculate (or at least estimate) probabilities of events give you?

Well, probability theory is really about providing a measure for our uncertainty about an event’s occurrence and/or giving us insights about the frequency of an event’s long-term occurrence. In short, it helps us build good expectations about real-world events and phenomena. And, consequently, this helps us make better decisions (in the most general sense).

There’s uncertainty in so many fields. You can apply probability theory in science, games, economics, education, politics, and many more. Really, it’s hard to even come up with examples where probability theory can’t help. Regardless of what you do or find interesting, probability theory is a very useful tool to have under your belt.

Well, that’s how I feel about it anyway.

My motivation for these posts

Convincing you that probability theory is cool.

So, in an effort to justify my position, in this series I want to show you many probability questions from diverse areas in life. I’m going to start with simpler problems which are more fun than useful. And, eventually, I’m going to build up to more complicated ones.

More importantly, the process of solving these problems itself is useful in training your brain to think about probability questions. Often, the principles involved in solving simpler problems are the same (or at least similar) to the ones used for solving more complex ones.

Even though most of my posts so far have been theoretical, I’ve also written a few more practical ones. For example, I’ve shown you how to apply some of the theoretical concepts from the beginning for things like:

• solving the inverse problem
• calculating the bias of a coin
• predicting presidential elections
• cryptography
• Occam’s razor

But in more than one occasion I’ve been asked to give more examples of practical applications of the theoretical concepts, as well as just examples of solving probability related problems. Hopefully, this series will be a good first step in this direction.

Probability questions from the book Understanding Probability

People have also asked me for recommendations on probability theory and statistics books that give a decent overview of all important concepts from these fields.

For the first posts in this series, I’m going to use twelve probability questions from the book Understanding Probability: Chance Rules in Everyday Life by the author Henk Tijms . Tijms is a Dutch mathematician who specializes in probability theory and many related fields. If you’ve been interested in probability theory for long enough, this is a name that you’ve likely already heard.

I personally read this book a little less than 10 years ago while I was still finishing my master’s degree in cognitive neuroscience and back then I found it one of the most interesting books on the subject. I was pleasantly surprised when I recently received an email from Henk Tijms himself in which he shared some positive words about Probabilistic World. And he was kind enough to give me permission to use the probability questions from his book in my posts.

The very first image in this post (the funny laundry cartoon) is actually from the same book. It is the header image of the first chapter in which the twelve questions are introduced. I say introduced because the actual solutions are given in later chapters.

Anyway, if you’re new to probability theory and statistics and you’re looking for a good comprehensive book on the subject, I recommend you start with this book. Now, some of the concepts Henk Tijms discusses in the book are things that I’ve discussed myself. And the rest are things I’m going to discuss in the future. But when you read about the same concept explained in different ways by different people, this helps you consolidate your knowledge and understanding. This is an approach that I myself have used for a very long time and I find it very effective in learning.

So, I think the Understanding Probability book is a very good complement to my website.

Answering probability questions with simulations

My third main motive for this series is that I want to introduce you to the method of answering probability questions using simulations. This is an extremely important technique and sometimes it’s the only way certain questions can be answered. Why? Well, as you’ll see in future posts, there are a lot of problems for which we don’t have an analytic solution.

I’ve already used simulations in some of my previous posts:

• estimating coin bias
• the mean, the mode, and the median
• the law of large numbers
• expected value
• mean and variance of probability distributions

But, except for the first post in this list, I didn’t share the computer code used in these simulations. In order to show you how to use simulations yourself, in this series I’m going to be much more explicit with my explanations. And, for all simulations, I’m going to use my favorite programming language Python .

Python is an extremely powerful language and is one of the top choices for programmers, scientists, and basically anybody doing math-related programming for whatever reason. It’s also extremely beginner-friendly, easy to learn, and surprisingly similar to a natural language (English).

But don’t worry. If you don’t know anything about Python or programming in general, I’m going to make sure you still benefit from this series to the fullest extent. The simulations themselves are going to be ones you can perform even with a pen and paper. The role of the programming code is simply to make your computer perform the same steps automatically and much faster. Even without a programming background, you’ll gain intuition about the simulations.

For each probability question, I’m going to first show its analytic solution and then compare it to the answer we get with a simulation. Meaning, we’re going to reach the same answer from two entirely different paths. Which is going to be a very useful exercise for gaining intuition about the law of large numbers too!

What is a computer simulation?

In a nutshell, computer simulations are used for estimating probabilities empirically. This involves repeating the process that leads to the outcomes we’re interested in a large number of times. In the meantime, you simply keep track of the number of times each outcome occur. And the goal of the computer is to automate the steps in order to save you (lots of) time and effort.

In my post on the law of large numbers I showed you a few examples of such empirical estimates of probabilities. When it comes to the process of flipping a fair coin, the law guarantees that the percentage of flips that turn up “heads” will converge to the probability of “heads”. Click on the image below to see how the empirical estimate of the probability converges to the real probability as the number of simulated flips increases:

Click on the image to start/restart the animation.

Technically, you don’t need a computer for this. Just take a real coin or a real die and flip/roll it multiple times while keeping track of the outcomes. Of course, doing it like that is just extremely laborious (especially for more complicated processes), so it’s much better to use a computer simulation.

Bottom line is that, as long as you can simulate the outcome generating process with a computer (with programming or otherwise), you can empirically estimate the probability of any outcome. All thanks to the law of large numbers!

You know nothing about programming?

If you’ve never done any programming in your life but still want to run the simulations, you can do it. And I really mean that. You don’t have to first read a book about programming or Python. You don’t have to follow any online tutorials. None of that.

Don’t get me wrong, if you’re generally interested in getting into programming, you can do those things as well. But I’m a big fan of the philosophy called “learning by doing”. Especially for programming. If right now you’re thinking to yourself “Really? I can still run and understand the code even if know absolutely nothing about programming?”… Yes, trust me, you will be able to. And you’ll most likely start picking up programming concepts in the process, even if you don’t set this as an explicit goal.

For one thing, you’ll be able to run the code by simple copy/pasting even if you don’t understand it at all. But, like I said, Python is one of the most readable programming languages in existence and, even if you read the code as if you were reading plain English, you’ll still understand a lot. Especially combined with my brief explanations.

By the way, like I said earlier, even if you choose to skip the programming parts of my posts, you won’t lose anything from the analytic answers to the probability questions. But if you want to make your first steps in programming with actual probability questions, this is going to be a very good opportunity for you. And the only thing you’re going to need to start is Python itself.

Normally, you can simply download and install Python from the official Python website . But if you’re completely new to Python and/or programming, I strongly recommend installing it with the platform called Anaconda and using it with the web application called Jupyter Notebook that comes along with Anaconda.

Anaconda and Jupyter Notebook

You can download Anaconda from their official website . Definitely download the one with the latest Python 3 version (not Python 2) and just be careful to choose the right option for your operating system.

Once you download and install Anaconda, you can get familiar with it by following this quick guide . In particular, pay attention to the part about Jupyter Notebook . This is an awesome web application for running Python code (among other things) that is automatically installed when you install Anaconda.

Jupyter Notebook is an extremely popular tool among programmers working in fields like data science, machine learning, artificial intelligence, and many others intersecting with probability theory, statistics, and mathematics in general. It runs in your browser (the same one you’re currently reading this post from) and you’ll be able to run the code from my posts with it. Not only that, Anaconda will automatically install many popular Python packages that have a ton of useful functionality for the fields I mentioned.

If you want to get your hands dirty, you can take a look at this somewhat more extensive Jupyter Notebook tutorial . But you don’t have to read all these things at once, you can also do that when you start practicing with the code from my posts.

Bottom line, all you need to do to be able to start running the code form my posts is:

• Download and install Anaconda
• Learn how to run Jupyter Notebook from the command prompt (spoiler: the command is simply jupyter notebook )
• Optionally, go through the short tutorials I linked to

If you encounter any issues with these steps, let me know in the comments below and me or another reader will help you what that.

The probability questions

So, here are the titles of the twelve probability questions, as listed in the opening chapter of the book Understanding Probability:

• A birthday problem ( analytic solution and Python simulation )
• Probability of winning streaks
• A scratch-and-win lottery
• A lotto problem
• Hitting the jackpot
• Who is the murderer?
• A coincidence problem
• A sock problem
• A statistical test problem
• The best-choice problem
• The Monty Hall dilemma
• An offer you can’t refuse — or can you?

Many of these questions are famous problems in probability theory. But here Henk Tijms presents them in a fun and informal format. My posts won’t necessarily be in the same order, since answering some of these questions requires knowledge of concepts I haven’t talked about yet and I might put them on hold until I do.

Of course, these twelve questions are only a starting point. I’m going to write many other posts on other questions, some of them famous, some of them ones I came up with myself. And yet others which are simply interesting real-world questions that can be answered with probability theory.

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How to Solve Probability Problems? (+FREE Worksheet!)

Do you want to know how to solve Probability Problems? Here you learn how to solve probability word problems.

Related Topics

• How to Interpret Histogram
• How to Interpret Pie Graphs
• How to Solve Permutations and Combinations
• How to Find Mean, Median, Mode, and Range of the Given Data

Step by step guide to solve Probability Problems

• Probability is the likelihood of something happening in the future. It is expressed as a number between zero (can never happen) to \(1\) (will always happen).
• Probability can be expressed as a fraction, a decimal, or a percent.
• To solve a probability problem identify the event, find the number of outcomes of the event, then use probability law: \(\frac{number\ of \ favorable \ outcome}{total \ number \ of \ possible \ outcomes}\)

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Probability problems – example 1:.

If there are \(8\) red balls and \(12\) blue balls in a basket, what is the probability that John will pick out a red ball from the basket?

There are \(8\) red balls and \(20\) a total number of balls. Therefore, the probability that John will pick out a red ball from the basket is \(8\) out of \(20\) or \(\frac{8}{8+12}=\frac{8}{20}=\frac{2}{5}\).

Probability Problems – Example 2:

A bag contains \(18\) balls: two green, five black, eight blue, a brown, a red, and one white. If \(17\) balls are removed from the bag at random, what is the probability that a brown ball has been removed?

If \(17\) balls are removed from the bag at random, there will be one ball in the bag. The probability of choosing a brown ball is \(1\) out of \(18\). Therefore, the probability of not choosing a brown ball is \(17\) out of \(18\) and the probability of having not a brown ball after removing \(17\) balls is the same.

Exercises for Solving Probability Problems

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• A number is chosen at random from \(1\) to \(10\). Find the probability of selecting a \(4\) or smaller.
• A number is chosen at random from \(1\) to \(50\). Find the probability of selecting multiples of \(10\).
• A number is chosen at random from \(1\) to \(10\). Find the probability of selecting of \(4\) and factors of \(6\).
• A number is chosen at random from \(1\) to \(10\). Find the probability of selecting a multiple of \(3\).
• A number is chosen at random from \(1\) to \(50\). Find the probability of selecting prime numbers.
• A number is chosen at random from \(1\) to \(25\). Find the probability of not selecting a composite number.

Download Probability Problems Worksheet

• \(\color{blue}{\frac{2}{5}}\)
• \(\color{blue}{\frac{1}{10}}\)
• \(\color{blue}{\frac{1}{2}}\)
• \(\color{blue}{\frac{3}{10}}\)
• \(\color{blue}{\frac{9}{25}}\)

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Probability

Probability  means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen. The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the  probability distribution , where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.

Learn More here: Study Mathematics

Probability Definition in Math

Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are going to happen, using it. Probability can range from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event. Probability for Class 10 is an important topic for the students which explains all the basic concepts of this topic. The probability of all the events in a sample space adds up to 1.

For example , when we toss a coin, either we get Head OR Tail, only two possible outcomes are possible (H, T). But when two coins are tossed then there will be four possible outcomes,  i.e {(H, H), (H, T), (T, H),  (T, T)}.

Download this lesson as PDF: – Download PDF Here

Formula for Probability

The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.

Sometimes students get mistaken for “favourable outcome” with “desirable outcome”. This is the basic formula. But there are some more formulas for different situations or events.

Solved Examples

1) There are 6 pillows in a bed, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow pillow?

Ans: The probability is equal to the number of yellow pillows in the bed divided by the total number of pillows, i.e. 2/6 = 1/3.

2) There is a container full of coloured bottles, red, blue, green and orange. Some of the bottles are picked out and displaced. Sumit did this 1000 times and got the following results:

• No. of blue bottles picked out: 300
• No. of red bottles: 200
• No. of green bottles: 450
• No. of orange bottles: 50

a) What is the probability that Sumit will pick a green bottle?

Ans: For every 1000 bottles picked out, 450 are green.

Therefore, P(green) = 450/1000 = 0.45

b) If there are 100 bottles in the container, how many of them are likely to be green?

Ans: The experiment implies that 450 out of 1000 bottles are green.

Therefore, out of 100 bottles, 45 are green.

Probability Tree

The tree diagram helps to organize and visualize the different possible outcomes. Branches and ends of the tree are two main positions. Probability of each branch is written on the branch, whereas the ends are containing the final outcome. Tree diagrams are used to figure out when to multiply and when to add. You can see below a tree diagram for the coin:

Types of Probability

There are three major types of probabilities:

Theoretical Probability

Experimental probability, axiomatic probability.

It is based on the possible chances of something to happen. The theoretical probability is mainly based on the reasoning behind probability. For example, if a coin is tossed, the theoretical probability of getting a head will be ½.

It is based on the basis of the observations of an experiment. The experimental probability can be calculated based on the number of possible outcomes by the total number of trials. For example, if a coin is tossed 10 times and head is recorded 6 times then, the experimental probability for heads is 6/10 or, 3/5.

In axiomatic probability, a set of rules or axioms are set which applies to all types. These axioms are set by Kolmogorov and are known as Kolmogorov’s three axioms. With the axiomatic approach to probability, the chances of occurrence or non-occurrence of the events can be quantified. The axiomatic probability lesson covers this concept in detail with Kolmogorov’s three rules (axioms) along with various examples.

Conditional Probability is the likelihood of an event or outcome occurring based on the occurrence of a previous event or outcome.

Probability of an Event

Assume an event E can occur in r ways out of a sum of n probable or possible equally likely ways . Then the probability of happening of the event or its success is expressed as;

The probability that the event will not occur or known as its failure is expressed as:

P(E’) = (n-r)/n = 1-(r/n)

E’ represents that the event will not occur.

Therefore, now we can say;

P(E) + P(E’) = 1

This means that the total of all the probabilities in any random test or experiment is equal to 1.

What are Equally Likely Events?

When the events have the same theoretical probability of happening, then they are called equally likely events. The results of a sample space are called equally likely if all of them have the same probability of occurring. For example, if you throw a die, then the probability of getting 1 is 1/6. Similarly, the probability of getting all the numbers from 2,3,4,5 and 6, one at a time is 1/6. Hence, the following are some examples of equally likely events when throwing a die:

• Getting 3 and 5 on throwing a die
• Getting an even number and an odd number on a die
• Getting 1, 2 or 3 on rolling a die

are equally likely events, since the probabilities of each event are equal.

Complementary Events

The possibility that there will be only two outcomes which states that an event will occur or not. Like a person will come or not come to your house, getting a job or not getting a job, etc. are examples of complementary events. Basically, the complement of an event occurring in the exact opposite that the probability of it is not occurring. Some more examples are:

• It will rain or not rain today
• The student will pass the exam or not pass.
• You win the lottery or you don’t.

Also, read: 

• Independent Events
• Mutually Exclusive Events

Probability Theory

Probability theory had its root in the 16th century when J. Cardan, an Italian mathematician and physician, addressed the first work on the topic, The Book on Games of Chance. After its inception, the knowledge of probability has brought to the attention of great mathematicians. Thus, Probability theory is the branch of mathematics that deals with the possibility of the happening of events. Although there are many distinct probability interpretations, probability theory interprets the concept precisely by expressing it through a set of axioms or hypotheses. These hypotheses help form the probability in terms of a possibility space, which allows a measure holding values between 0 and 1. This is known as the probability measure, to a set of possible outcomes of the sample space.

Probability Density Function

The Probability Density Function (PDF) is the probability function which is represented for the density of a continuous random variable lying between a certain range of values. Probability Density Function explains the normal distribution and how mean and deviation exists. The standard normal distribution is used to create a database or statistics, which are often used in science to represent the real-valued variables, whose distribution is not known.

Probability Terms and Definition

Some of the important probability terms are discussed here:

Applications of Probability

Probability has a wide variety of applications in real life. Some of the common applications which we see in our everyday life while checking the results of the following events:

• Choosing a card from the deck of cards
• Flipping a coin
• Throwing a dice in the air
• Pulling a red ball out of a bucket of red and white balls
• Winning a lucky draw

Other Major Applications of Probability

• It is used for risk assessment and modelling in various industries
• Weather forecasting or prediction of weather changes
• Probability of a team winning in a sport based on players and strength of team
• In the share market, chances of getting the hike of share prices

Problems and Solutions on Probability

Question 1: Find the probability of ‘getting 3 on rolling a die’.

Sample Space = S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = n(S) = 6

Let A be the event of getting 3.

Number of favourable outcomes = n(A) = 1

i.e. A  = {3}

Probability, P(A) = n(A)/n(S) = 1/6

Hence, P(getting 3 on rolling a die) = 1/6

Question 2: Draw a random card from a pack of cards. What is the probability that the card drawn is a face card?

A standard deck has 52 cards.

Total number of outcomes = n(S) = 52

Let E be the event of drawing a face card.

Number of favourable events = n(E) = 4 x 3 = 12 (considered Jack, Queen and King only)

Probability, P = Number of Favourable Outcomes/Total Number of Outcomes

P(E) = n(E)/n(S)

P(the card drawn is a face card) = 3/13

Question 3: A vessel contains 4 blue balls, 5 red balls and 11 white balls. If three balls are drawn from the vessel at random, what is the probability that the first ball is red, the second ball is blue, and the third ball is white?

The probability to get the first ball is red or the first event is 5/20.

Since we have drawn a ball for the first event to occur, then the number of possibilities left for the second event to occur is 20 – 1 = 19.

Hence, the probability of getting the second ball as blue or the second event is 4/19.

Again with the first and second event occurring, the number of possibilities left for the third event to occur is 19 – 1 = 18.

And the probability of the third ball is white or the third event is 11/18.

Therefore, the probability is 5/20 x 4/19 x 11/18 = 44/1368 = 0.032.

Or we can express it as: P = 3.2%.

Question 4: Two dice are rolled, find the probability that the sum is:

• less than 13

Video Lectures

Introduction.

Solving Probability Questions

Probability Important Topics

Probability Important Questions

Probability Problems

• Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is: (i) 6 (ii) 12 (iii) 7
• A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a (i) red ball (ii) green ball (iii) not a blue ball
• All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value (i) 7 (ii) greater than 7 (iii) less than 7
• A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded. (i) How many different scores are possible? (ii) What is the probability of getting a total of 7?

Frequently Asked Questions (FAQs) on Probability

What is probability give an example, what is the formula of probability, what are the different types of probability, what are the basic rules of probability, what is the complement rule in probability.

In probability, the complement rule states that “the sum of probabilities of an event and its complement should be equal to 1”. If A is an event, then the complement rule is given as: P(A) + P(A’) = 1.

What are the different ways to present the probability value?

The three ways to present the probability values are:

• Decimal or fraction

What does the probability of 0 represent?

The probability of 0 represents that the event will not happen or that it is an impossible event.

What is the sample space for tossing two coins?

The sample space for tossing two coins is: S = {HH, HT, TH, TT}

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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good basics concepts provide

Great video content. Perfectly explained for examinations. I really like to learn from BYJU’s

Thank you for your best information on probablity

Good explanation about probability and concept for simple understanding the overall chapter. I hope this is a good way to understand the CONCEPT

For learner of class X standard , it is providing all the relevant informations and approach towards the contenet is knitted in an elegant manner and students will have the opportunity to grasp the topic easily and will be immensely benefited.

this topic has been hard for me but now I know what it is all about and I have really enjoyed it thanks for your good explanation

Well explained

Excellent explanation of probability. One can easily understand about the probability.

I really appreciated your explanations because it’s well understandable Thanks

Love the way you teach

There are 3 boxes Box A contains 10 bulbs out of which 4 are dead box b contains 6 bulbs out of which 1 is dead box c contains 8 bulbs out of which 3 are dead. If a dead bulb is picked at random find the probability that it is from which box?

Probability of selecting a dead bulb from the first box = (1/3) x (4/10) = 4/30 Probability of selecting a dead bulb from the second box = (1/3) x (1/6) = 1/18 Probability of selecting a dead bulb from the third box = (1/3) x (3/8) = 3/24 = 1/8 Total probability = (4/30) + (1/18) + (1/8) = (48 + 20 + 45)360 =113/360

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Probability Problems: How to Solve Probability Problems the Easy Way!

Probability and Statistics > Probability Index > Probability problems

Probability Problems Overview.

If you’re taking a class in basic probability, right now you’re probably feeling utterly confused with the rules of probability. After all, there are lots . When do add? When to multiply? When to use combinations ? Ugh. I remember when I was learning probability and even after I passed the class I still struggled with the heads and tails of figuring out when to use what rule. Here’s a summary of common situations which will show you how to solve probability problems using the right technique.

The types of probability problems shown here are simple events , like the odds of choosing something or winning something. Later on in probability, you’ll be coming across probability distributions like the binomial distribution and the normal distribution . You’ll usually know you’re solving a probability distribution problem by keywords like “normally distributed” or “fits a binomial distribution.” If that’s the case, you’ll want to check out the probability index for more articles on probability problems that involve distributions.

Click on the description below that describes the type of probability problem you have:

Probability Problems about Events.

Finding the probability of a simple event happening is fairly straightforward: add the probabilities together . For example, if you have a 10% chance of winning $10 and a 25% chance of winning $20 then your overall odds of winning something is 10% + 25% = 35%. This only works for mutually exclusive events (events that cannot happen at the same time).

Dice rolling probability problems.

To solve dice rolling problems, you could have one dice, or you could have three dice. The probability will change depending on how many dice you are rolling and what numbers you want to pick. The easiest way to solve these types of probability problems is to write out all the possible dice combinations (that’s called writing a sample space ). A very simple example, if you want to know the probability of rolling a double with two die, your sample space would be:

[1][1], [1][2], [1][3], [1][4], [1][5], [1][6], [2][1], [2][2], [2][3], [2][4],[2][5], [2][6], [3][1], [3][2], [3][3], [3][4], [3][5], [3][6], [4][1], [4][2], [4][3], [4][4], [4][5], [4][6], [5][1], [5][2], [5][3], [5][4], [5][5], [5][6], [6][1], [6][2], [6][3], [6][4], [6][5], [6][6].

There are six doubles: [1][1], [2][2], [3][3], [4][4], [5][5], [6][6] and 36 possible rolls, so the probability is 6/36. You could use the same sample space to figure out your odds of rolling a 3 and a 4 (2/36) or that the two die add up to 7. In the last case, there’s a [6][1], [1][6], [3][4], [4][3], [5,2], [2,5] so the probability is 6/36.

For more, see: Dice Rolling Probability .

Card probability problems.

You can use the same technique that’s used for dice rolling (see above): Write out your sample space. For one standard deck of cards, you have 52 cards. Your sample space is:

hearts: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A clubs: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A spades: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A diamonds: 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A

If you were asked the probability of choosing a spade or a 2, there are 13 spades (including the 2 of spades) and three other “2”s, making 16 cards. So your probability is 16/52.

For more, see: Probability of picking from a deck of cards .

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Use These Examples of Probability To Guide You Through Calculating the Probability of Simple Events

Probability is the chance or likelihood that an event will happen.

It is the ratio of the number of ways an event can occur to the number of possible outcomes. We'll use the following model to help calculate the probability of simple events.

As you can see, with this formula, we will write the probability of an event as a fraction.

The numerator (in red) is the number of chances and the denominator (in blue) is the set of all possible outcomes. This is also known as the sample space.

Let's take a look at a few examples of probability.

Example 1- Probability Using a Die

Given a standard die, determine the probability for the following events when rolling the die one time:

P(even number)

Before we start the solution, please take note that:

P(5) means the probability of rolling a 5

When you see P(  ) this means to find the probability of whatever is indicated inside of the parenthesis.

Solutions :

Let’s first identify the sample space.  The sample space then becomes the denominator in our fraction when calculating probability.

Sample Space :      6       We are using a standard die.  A standard die has 6 sides and contains the numbers 1-6. 

Therefore, our sample space is 6 because there are 6 total outcomes that could occur when we roll the die.  The 6 outcomes are: 1, 2, 3, 4, 5, 6

Special Note:

Always simplify your fraction if possible!

Now let's take a look at a probability situation that involves marbles.

Example 2 - Probability with Marbles

There are 4 blue marbles, 5 red marbles, 1 green marble, and 2 black marbles in a bag.  Suppose you select one marble at random.  Find each probability.

P(blue or black)

P(not green)

P(not purple)

Hopefully these two examples have helped you to apply the formula in order to calculate the probability for any simple event.

Now, it's your turn to try! Check out the spinner in the practice problem below.

Practice Problem

Great Job! You've got the basics, now you are ready to move on to the next lesson on Tree Diagrams & The Fundamental Counting Principle.

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• Right Triangle

1. Define and characterize probability.

• number of the events is finite
• all events have the same chance to occur
• no two events can occur in the same time

2. There are 18 tickets marked with numbers 1 to 18. What‘s the probability of selecting a ticket having the following property:

3. Determine the probability of following results when throwing 2 playing cubes (a red one and a blue one):

4. A gambler playing with 3 playing cubes wants to know weather to bet on sum 11 or 12. Which of the sums will occur more probably?

5. 82 170 of 100 000 children live 40 years and 37 930 of 100 000 children live 70 years. Determine the probability of a 40 years old person to live 70 years.

6. In a town there are 4 crossroads with trafic lights. Each trafic light opens or closes the traffic with the same probability of 0.5. Determine the probability of:

7. 32 playing cards include 4 aces and 12 figures. Determine the probability of a randomly selected card to be an ace or a figure.

Solution: 

8. Determine the probability of 3 of 5 born children being sons if the probability of a children to be a boy equals P(A) = 0,51.

9. There are 16 cola bottles on the table. 10 of them are filled by Coca Cola and 6 of them are filled by Pepsi. Determine the probability of 4 randomly selected bottles to include 2 Coca Cola and 2 Pepsi bottles.

10. In a game of chance 6 of 49 numbers are the winning numbers. Determine the probability of reaching:

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Free Mathematics Tutorials

Bayer's theorem examples with solutions.

Bayes' theorem to find conditional porbabilities is explained and used to solve examples including detailed explanations. Diagrams are used to give a visual explanation to the theorem. Also the numerical results obtained are discussed in order to understand the possible applications of the theorem.

Bayes' theorem

Use of Bayes' Thereom Examples with Detailed Solutions

Example 3 Three factories produce light bulbs to supply the market. Factory A produces 20%, 50% of the tools are produced in factories B and 30% in factory C. 2% of the bulbs produced in factory A, 1% of the bulbs produced in factory B and 3% of the bulbs produced in factory C are defective. A bulb is selected at random in the market and found to be defective. what is the probability that this bulb was produced by factory B? Solution to Example 3 Let \( P(A) = 20\% \), \( P(B) = 50\% \) and \( P(C) = 30\% \) represent the probabilities that a bulb selected at random is from factory A, B and C respectively. Let \( P(D) \) be the probability that a defective bulb is selected. Let \( P(D | A) = 2\% \), \( P(D | B) = 1\% \) and \( P(D | C) = 3\%\) represent the conditional probabilities that a bulb is defective given that it is selected from factory A, B and C respectively. We now calculate that the conditional probability that the bulb was produced by factory B given that it is defective written as \( P(B | D) \) and given by Bayes' theorem. \( P(B | D) = \dfrac{P(D | B) P(B) }{ P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C)}\) \( = \dfrac{1\% \times 50\%}{ 2\% \times 20\% + 1\% \times 50\% + 3\% \times 30\%} = 0.2777\) Although factory B produces 50% of the bulbs, the probability that the selected (defective) bulb comes from this factory is low because the bulbs produced by this factory have low probability (1%) of being defective.

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One of the best ways to learn statistics is to solve practice problems. These problems test your understanding of statistics terminology and your ability to solve common statistics problems. Each problem includes a step-by-step explanation of the solution.

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In one state, 52% of the voters are Republicans, and 48% are Democrats. In a second state, 47% of the voters are Republicans, and 53% are Democrats. Suppose a simple random sample of 100 voters are surveyed from each state.

What is the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state?

The correct answer is C. For this analysis, let P 1 = the proportion of Republican voters in the first state, P 2 = the proportion of Republican voters in the second state, p 1 = the proportion of Republican voters in the sample from the first state, and p 2 = the proportion of Republican voters in the sample from the second state. The number of voters sampled from the first state (n 1 ) = 100, and the number of voters sampled from the second state (n 2 ) = 100.

The solution involves four steps.

• Make sure the sample size is big enough to model differences with a normal population. Because n 1 P 1 = 100 * 0.52 = 52, n 1 (1 - P 1 ) = 100 * 0.48 = 48, n 2 P 2 = 100 * 0.47 = 47, and n 2 (1 - P 2 ) = 100 * 0.53 = 53 are each greater than 10, the sample size is large enough.
• Find the mean of the difference in sample proportions: E(p 1 - p 2 ) = P 1 - P 2 = 0.52 - 0.47 = 0.05.

σ d = sqrt{ [ P1( 1 - P 1 ) / n 1 ] + [ P 2 (1 - P 2 ) / n 2 ] }

σ d = sqrt{ [ (0.52)(0.48) / 100 ] + [ (0.47)(0.53) / 100 ] }

σ d = sqrt (0.002496 + 0.002491) = sqrt(0.004987) = 0.0706

z p 1 - p 2 = (x - μ p 1 - p 2 ) / σ d = (0 - 0.05)/0.0706 = -0.7082

Using Stat Trek's Normal Distribution Calculator , we find that the probability of a z-score being -0.7082 or less is 0.24.

Therefore, the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state is 0.24.

See also: Difference Between Proportions

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2.1.5 Solved Problems: Combinatorics

Let $A$ and $B$ be two finite sets, with $|A|=m$ and $|B|=n$. How many distinct functions (mappings) can you define from set $A$ to set $B$, $f:A \rightarrow B$?

• We can solve this problem using the multiplication principle. Let $$A =\{a_1,a_2,a_3,...,a_m\},$$ $$B =\{b_1,b_2,b_3,...,b_n\}.$$ Note that to define a mapping from $A$ to $B$, we have $n$ options for $f(a_1)$, i.e., $f(a_1) \in B=\{b_1,b_2,b_3,...,b_n\}$. Similarly we have $n$ options for $f(a_2)$, and so on. Thus by the multiplication principle, the total number of distinct functions $f:A \rightarrow B$ is $$n \cdot n \cdot n \cdots n = n^m.$$

A function is said to be one-to-one if for all $x_1\neq x_2$, we have $f(x_1)\neq f(x_2)$. Equivalently, we can say a function is one-to-one if whenever $f(x_1)=f(x_2)$, then $x_1=x_2$. Let $A$ and $B$ be two finite sets, with $|A|=m$ and $|B|=n$. How many distinct one-to-one functions (mappings) can you define from set $A$ to set $B$, $f:A \rightarrow B$?

• Again let $$A =\{a_1,a_2,a_3,...,a_m\},$$ $$B =\{b_1,b_2,b_3,...,b_n\}.$$ To define a one-to-one mapping from $A$ to $B$, we have $n$ options for $f(a_1)$, i.e., $f(a_1) \in B=\{b_1,b_2,b_3,...,b_n\}$. Given $f(a_1)$, we have $n-1$ options for $f(a_2)$, and so on. Thus by the multiplication principle, the total number of distinct functions $f:A \rightarrow B$, is $$n \cdot (n-1) \cdot (n-2) \cdots (n-m+1) = P^n_{m}.$$ Thus, in other words, choosing a one-to-one function from $A$ to $B$ is equivalent to choosing an $m$-permutation from the $n$-element set $B$ (ordered sampling without replacement) and as we have seen there are $P^n_{m}$ ways to do that.

An urn contains $30$ red balls and $70$ green balls. What is the probability of getting exactly $k$ red balls in a sample of size $20$ if the sampling is done with replacement (repetition allowed)? Assume $0\leq k \leq 20$.

• Here any time we take a sample from the urn we put it back before the next sample (sampling with replacement). Thus in this experiment each time we sample, the probability of choosing a red ball is $\frac{30}{100}$, and we repeat this in $20$ independent trials. This is exactly the binomial experiment. Thus, using the binomial formula we obtain $$P(k \textrm{ red balls})={20 \choose k} (0.3)^k(0.7)^{20-k}.$$

An urn consists of $30$ red balls and $70$ green balls. What is the probability of getting exactly $k$ red balls in a sample of size $20$ if the sampling is done without replacement (repetition not allowed)?

• Let $A$ be the event (set) of getting exactly $k$ red balls. To find $P(A)=\frac{|A|}{|S|}$, we need to find $|A|$ and $|S|$. First, note that $|S|={100 \choose 20}$. Next, to find $|A|$, we need to find out in how many ways we can choose $k$ red balls and $20-k$ green balls. Using the multiplication principle, we have $$|A|= {30 \choose k}{70 \choose 20-k}.$$ Thus, we have $$P(A)= \frac{{30 \choose k}{70 \choose 20-k}}{{100 \choose 20}}.$$
• $k=5$ with probability $\frac{1}{4}$;
• $k=10$ with probability $\frac{1}{4}$;
• $k=15$ with probability $\frac{1}{2}$.
• What is the probability that at least two of them have been born in the same month? Assume that all months are equally likely.
• Given that we already know there are at least two people that celebrate their birthday in the same month, what is the probability that $k=10$?

How many distinct solutions does the following equation have? $$x_1+x_2+x_3+x_4=100, \textrm{ such that }$$ $$x_1 \in \{1,2,3..\}, x_2 \in \{2,3,4,..\}, x_3,x_4 \in \{0,1,2,3,...\}.$$

• We already know that in general the number of solutions to the equation $$x_1+x_2+...+x_n=k, \textrm{ where } x_i \in \{0,1,2,3,...\}$$ is equal to $${n+k-1 \choose k}={n+k-1 \choose n-1}.$$ We need to convert the restrictions in this problem to match this general form. We are given that $x_1 \in \{1,2,3..\}$, so if we define $$y_1=x_1-1,$$ then $y_1 \in \{0,1,2,3,...\}$. Similarly define $y_2 =x_2-2$, so $y_2 \in \{0,1,2,3,...\}$. Now the question becomes equivalent to finding the number of solutions to the equation $$y_1+1+y_2+2+x_3+x_4=100, \textrm{ where } y_1,y_2,x_3,x_4 \in \{0,1,2,3,...\},$$ or equivalently, the number of solutions to the equation $$y_1+y_2+x_3+x_4=97, \textrm{ where } y_1,y_2,x_3,x_4 \in \{0,1,2,3,...\}.$$ As we know, this is equal to $${4+97-1 \choose 3}={100 \choose 3}.$$

Here is a famous problem: $N$ guests arrive at a party. Each person is wearing a hat. We collect all hats and then randomly redistribute the hats, giving each person one of the $N$ hats randomly. What is the probability that at least one person receives his/her own hat? Hint: Use the inclusion-exclusion principle.

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Quantum annealing showing an exponentially small success probability despite a constant energy gap with polynomial energy

Hiroshi hayasaka, takashi imoto, yuichiro matsuzaki, and shiro kawabata, phys. rev. a 110 , 022620 – published 26 august 2024.

• No Citing Articles
• INTRODUCTION
• QUANTUM ANNEALING AND ADIABATIC GROVER…
• GENERAL FRAMEWORK
• ADIABATIC GROVER SEARCH WITH PENALTY…
• FERROMAGNETIC p SPIN MODEL
• ACKNOWLEDGMENTS

Quantum annealing (QA) is a method for solving combinatorial optimization problems. We can estimate the computational time for QA using what is referred to as the adiabatic condition derived from the adiabatic theorem. The adiabatic condition consists of two parts: an energy gap and a transition matrix. Most past studies have focused on the relationship between the energy gap and computational time. The success probability of QA is considered to decrease exponentially owing to the exponentially decreasing energy gap at the first-order phase-transition point. In this study, through a detailed analysis of the relationship between the energy gap, transition matrix, and computational cost during QA, we propose a general method for constructing counterintuitive models wherein QA with a constant annealing time fails despite a constant energy gap, based on polynomial energy. We assume that the energy of the total Hamiltonian is at most Θ ( L ) , where L is the number of qubits. In our formalism, we choose a known model that exhibits an exponentially small energy gap during QA, and modify the model by adding a specific penalty term to the Hamiltonian. In the modified model, the transition matrix in the adiabatic condition becomes exponentially large as the number of qubits increases, while the energy gap remains constant. Moreover, we achieve a quadratic speedup, for which the upper bound for improvement in the adiabatic condition is determined by the polynomial energy. For concrete examples, we consider the adiabatic Grover search and the ferromagnetic p -spin model. In these cases, with the addition of the penalty term, although the success probability of QA on the modified models becomes exponentially small despite a constant energy gap, we are able to achieve a success probability considerably higher than that of conventional QA. Moreover, in concrete examples, we numerically show the scaling of the computational cost is quadratically improved compared to the conventional QA. Our findings pave the way for a better understanding of QA performance.

• Received 17 November 2023
• Revised 11 July 2024
• Accepted 5 August 2024

DOI: https://doi.org/10.1103/PhysRevA.110.022620

©2024 American Physical Society

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Authors & Affiliations

• 1 Global Research and Development Center for Business by Quantum-AI Technology (G-QuAT), National Institute of Advanced Industrial Science and Technology (AIST), 1-1-1, Umezono, Tsukuba, Ibaraki 305-8568, Japan
• 2 NEC-AIST Quantum Technology Cooperative Research Laboratory, National Institute of Advanced Industrial Science and Technology (AIST), 1-1-1, Umezono, Tsukuba, Ibaraki 305-8568, Japan
• * Contact author: [email protected]
• † Contact author: [email protected]
• ‡ Contact author: [email protected], present address: [email protected]
• § Contact author: [email protected], present address: [email protected]

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Vol. 110, Iss. 2 — August 2024

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Plot of adiabatic-condition term η for the adiabatic Grover search as a function of normalized time t / T , where L is the number of qubits.

Energy spectrum and fidelity for the adiabatic Grover search as functions of t / T . Energy diagram in L = 10 (a) without penalty term and (b) with penalty term. Fidelity (c) without penalty term and (d) with penalty term. Annealing time T = 20 .

Scaling of (a) energy gap at t = T / 2 and (b) fidelity at t = T for the adiabatic Grover search. Purple (bottom) and green (top) lines correspond to cases without and with penalty term, respectively. In (a) and (b), annealing time is T = 20 . (c) The fidelity and corresponding computational cost where the system size is L = 10 . Here, we use T = 10 n ( T = 100 m ) for 1 ≤ n ≤ 20 , n ∈ N ( 1 ≤ m ≤ 60 , m ∈ N ) in the case with (without) penalty term. (d) The computational cost to obtain a fidelity higher than 0.5. We plot the cost against the number of qubits to identify the scaling behavior. In (c) and (d), purple (square) and green (circle) dots correspond to cases without and with penalty term, respectively.

Energy spectrum and fidelity as functions of t / T for the ferromagnetic p -spin model. Energy diagram in L = 40 (a) without penalty term and (b) with penalty term. Fidelity (c) without penalty term and (d) with penalty term. Annealing time T = 20 , p = 5 .

Scaling of (a) minimum of energy gap and (b) fidelity at t = T for the ferromagnetic p -spin model. Purple (bottom) and green (top) lines correspond to cases without and with penalty term, respectively. In (a) and (b), we set T = 20 and p = 5 . (c) The fidelity and corresponding computational cost where the system size is L = 16 . Here, we use T = 10 n ( T = 100 m ) for 1 ≤ n ≤ 20 , n ∈ N ( 1 ≤ m ≤ 20 , m ∈ N ) in the case with (without) penalty term. (d) The computational cost to obtain a fidelity higher than 0.5. We plot the cost against the number of qubits to identify the scaling behavior. In (c) and (d), purple (square) and green (circle) dots correspond to cases without and with penalty term, respectively.

Fidelity (a) without penalty term and (b) with penalty term as functions of t / T . Annealing time T = 20 .

Scaling of fidelity at t = T . Purple (bottom) and green (top) lines denote the cases without and with penalty term, respectively. Annealing time T = 20 .

Fidelity for the adiabatic Grover search with penalty term as functions of T . The system size is L = 10 .

Maximum of transition matrix for the ferromagnetic p -spin model. Purple (middle), green (top), and blue (bottom) lines denote cases without and with penalty term and with nonstoquastic term, respectively. Annealing time T = 20 .

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Using an Inquiry Process to Solve Persistent Classroom Problems

Teachers can resolve challenges that come up over and over by using data to keep testing strategies until they find what works.

The start of a new school year is always filled with anticipation. Teachers hope for engaged students who want to attain success. Students set personal goals and often hope that this year will be better than last year. Parents want their children to try hard, do well, and they want their children’s teachers to be supportive and offer a safe learning space. The new school year is often filled with hope.

However, despite all the best intentions, at some point, the teacher will encounter a problem. Many problems can be resolved with the knowledge acquired through a teacher’s experience. Students may forget to bring a pencil to class, so you just keep a jar of sharpened pencils on your desk. Students who are English language learners struggle to read Shakespeare, so you provide them and all the other students with a link to the audio version of the play that they can listen to. These impromptu decisions have the potential to swiftly address the problem, thereby eliminating the need for further investigation.

What is Teacher Inquiry?

But what is a teacher supposed to do if a problem persists over time? Some students are always late to class right after lunch. Some students never raise their hand to participate in a class discussion. Some students don’t effectively edit their work prior to handing it in. How can a teacher work to identify strategies that can solve these persistent classroom problems? This is where teacher inquiry becomes a valuable tool.

As Marilyn Cochran-Smith and Susan L. Lytle discuss in their book Inquiry as Stance: Practitioner Research for the Next Generation , teacher inquiry is a process of questioning, exploring, and implementing strategies to address persistent classroom challenges. It mirrors the active learning process we encourage in students and can transform recurring problems into opportunities for growth. Most important, it also creates space for students to share their voices and perspectives—allowing them to play a role in guiding the changes that are implemented in the classroom.

How to Start the Inquiry Process

Identify the problem. Begin by clearly defining the issue. For example, if students are frequently late after lunch, consider this as your inquiry focus.

Gather action information. Before rushing to solutions, gather insights from blogs, research, books, or colleagues. For instance, if the problem is tardiness, you might explore strategies like greeting students at the door or starting the class with a high-energy, collaborative activity that is engaging for students .

Frame your inquiry question. Craft a focused question using the format: What impact does X have on Y? Here X is the planned intervention, and Y is the behavior.

• What impact does greeting students at the classroom door have on their punctuality?
• What impact does a sharing circle have on students’ presentation anxiety?

This approach shifts the perspective from seeing students as the problem to exploring solutions to unwanted behaviors. Rather than saying, “Students are always late to class right after lunch,” we can ask, “What impact does an engaging collaborative activity at the start of class have on students’ punctuality?” 

Implementing and Assessing the Strategy

Plan data collection. Before implementing your strategy, decide how you’ll measure its effectiveness. This could involve the following:

Quantitative data: Use attendance records, test scores, or quick surveys—whether digital or paper-based—to track student engagement. For example, monitor the number of students arriving on time before and after you start greeting them. Choose the survey method that best fits your classroom’s needs, whether it’s a digital link or QR code for students with technology, or a paper survey for those without.

Qualitative data: Collect student feedback through informal interviews or reflective journals to understand their experiences.

Mixed methods: You can collect a combination of quantitative and qualitative data to allow for quick, easy-to-read facts (quantitative) with an understanding of the why (qualitative) for the data.

Tip: To avoid overwhelming yourself, use data that you’re already collecting and analyze it with your inquiry question in mind.

Implement the strategy. Start with a small, manageable change. If you’re trying to improve punctuality, greet students at the door for a week and note any changes.

Evaluating the Results

Analyze the data . Review your collected data to see if there’s a noticeable effect. Did more students arrive on time? If you used a survey, what do the results indicate about students’ attitudes?

Reflect on the outcome. If the strategy worked, consider how it can be sustained or adapted for other challenges. If it didn’t, reflect on why. Did the strategy need more time, or should a different approach be tried?

Example: If greeting students didn’t improve punctuality, consider if greeting needs to be combined with another intervention, like a change in seating arrangements or communicating with students’ families to remind them about the importance of punctuality.

What if the Strategy Doesn’t Work?

Not all inquiries lead to success, and that’s OK. If your initial strategy doesn’t yield the desired results, reflect on the process.

• What could be adjusted? Perhaps the data collection method wasn’t effective, or the strategy needs more time to show results.
• What did you learn? Even if the strategy didn’t solve the problem, what insights did you gain that could inform future inquiries?

Adopt the same growth mindset you encourage in your students in order to view setbacks as learning opportunities . Inquiry is a cycle of continuous improvement, not a onetime fix.

Embracing the Inquiry Mindset

Inquiry empowers teachers to approach challenges with curiosity and adaptability. By framing problems as opportunities to learn, gathering and analyzing data, and reflecting on outcomes, teachers model the persistence and growth mindset we aim to instill in our students. Even when results aren’t immediate, the process fosters a culture of continuous learning and improvement, benefiting teachers and students alike.

Designing a Context-Driven Problem-Solving Method with Metacognitive Scaffolding Experience Intervention for Biology Instruction

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• Published: 27 August 2024

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• Merga Dinssa Eticha   ORCID: orcid.org/0009-0008-9263-3273 1 , 2 ,
• Adula Bekele Hunde 3 &
• Tsige Ketema 1  

Learner-centered instructional practices, such as the metacognitive strategies scaffolding the problem-solving method for Biology instruction, have been shown to promote students’ autonomy and self-direction, significantly enhancing their understanding of scientific concepts. Thus, this study aimed to elucidate the importance and procedures of context analysis in the development of a context-driven problem-solving method with a metacognitive scaffolding instructional approach, which enhances students’ learning effectiveness in Biology. Therefore, the study was conducted in the Biology departments of secondary schools in Shambu Town, Oromia Region, Ethiopia. The study employed mixed-methods research to collect and analyze data, involving 12 teachers and 80 students. The data collection tools used were interviews, observations, and a questionnaire. The study revealed that conducting a context analysis that involves teachers, students, and learning contexts is essential in designing a context-driven problem-solving method with metacognitive scaffolding for Biology instruction, which provides authentic examples, instructional content, and engaging scenarios for teachers and students. As a result, the findings of this study provide a practical instructional strategy that can be applied to studies aimed at designing a context-driven problem-solving method with metacognitive scaffolding with the potential to influence instructional practices.

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Introduction

Biology is a vital subject in the Natural Sciences and enables learners to understand the mechanisms of living organisms and their practical applications for humans (Agaba, 2013 ). Therefore, Biology instruction requires interactive, learner-centered instructional methods like the problem-solving method with metacognitive scaffolding (PSMMS), which foster students to develop critical thinking, problem-solving, metacognitive, and scientific process skills (Al Azmy & Alebous, 2020 ; Inel & Balim, 2010 ) and help them make informed decisions regarding health and the environment, thereby advancing scientific knowledge (Aurah et al., 2011 ).

Although the focus is on students acquiring scientific knowledge and higher-order thinking skills (Senyigit, 2021 ), research revealed gaps in implementing the PSMMS in Biology, mainly due to the teachers’ limited experience in learner-centered methods (Agena, 2010 ; Beyessa, 2014 ), poor enhancement practices (MoE, 2019 ), tendency to use conventional problem-solving approaches (Aurah et al., 2011 ), and limited understanding of the roles of metacognition in instructional processes (Cimer, 2012 ). On the other hand, there is limited study on the importance of metacognitive instruction in scaffolding the problem-solving method in Biology, although it has a significant impact on students’ performance in mathematics and logical reasoning (Guner & Erbay, 2021 ).

In addition, metacognitive instructional strategies in primary school sciences and the contributions of metacognitive instructional intervention in developing countries are other areas where limited research has been done (Sbhatu, 2006). These challenges offer a study ground for investigating the intervention of metacognitive instructional methods in secondary schools, focusing on the problem-solving method in Biology. This study, therefore, aims to answer the research question, “How can context analysis be used to design a context-driven PSMMS and suggest PSMMS instructional guidelines to enhance students’ effective Biology learning?”

Theoretical Background

The problem-solving method.

The problem-solving method is a learner-centered approach that focuses on identifying, investigating, and solving problems (Ahmady & Nakhostin-Ruhi, 2014 ). The problem-solving method in Biology promotes advanced and critical thinking skills, enhancing students’ attitudes, academic performance, and subject understanding (Albay, 2019 ; Khaparde, 2019 ). Research has shown that students who learn using the problem-solving method outperform those who are taught conventionally (Nnorom, 2019 ). Studies have discussed that the problem-solving method encourages experimentation or learning through trial-and-error and also facilitates a constructivist learning environment by encouraging brainstorming and inquiry (e.g., Ishaku, 2015).

Metacognition

Metacognition, introduced by John Flavell in 1976, refers to an individual’s awareness, critical thinking, reflective judgment, and control of cognitive processes and strategies (Tachie, 2019 ). It consists of two main components, namely metacognitive knowledge and metacognitive regulation (Lai, 2011 ). Metacognitive knowledge involves understanding one’s own thinking, influencing performance, and effective use of methods through declarative, procedural, and conditional knowledge (Schraw et al., 2006 ; Sperling et al., 2004 ), while metacognitive regulation is about controlling thought processes and monitoring cognition, which involves planning, implementing, monitoring, and evaluating strategies (Aaltonen & Ikavalko, 2002 ; Zumbrunn et al., 2011 ).

Metacognitive instructional strategies are used to enhance learners’ effectiveness and support their learning process during the stages of forethought, performance, and self-reflection (Okoro & Chukwudi, 2011 ; Zimmerman, 2008 ). Therefore, metacognitive scaffolding, as described by Zimmerman ( 2008 ), is important in classroom interventions because it promotes problem-solving processes and supports metacognitive activities. According to Sbhatu (2006), understanding metacognitive processes and methods is fundamental for complex problem-solving tasks. Metacognitive functions are categorized based on the phases of the problem-solving method, including problem recognition, presentation, planning, execution, and evaluation (Kapa, 2001 ).

PSMMS in the Face of Globalization and Twenty-First Century Advancements

In the twenty-first century, societies rely on scientific and technological advances, and promoting scientific literacy is crucial for their integration into interactive learning environments (Chu et al., 2017 ). Studies suggest that science, technology, engineering, and mathematics (STEM) education promotes critical thinking, creativity, and problem-solving skills (Widya et al., 2019 ). Therefore, teachers should adopt a learning science and learner-centered approach and focus on higher-order thinking skills and problem-based tasks (Darling-Hammond et al., 2020 ; Nariman, 2014).

The implementation of metacognitive strategies as a scaffold system for the problem-solving method, which simultaneously fosters the development of higher-order skills in their Biology learning, helps students advance in the age of globalization and the twenty-first century. According to Chu et al. ( 2017 ), twenty-first century skills are classified into four categories, such as ways of thinking, ways of working, tools for working, and ways of living in an advanced world. Therefore, studies suggest that teachers can help students develop twenty-first century skills and influence learning through metacognition, thereby promoting self-directed learning (Stehle & Peters-Burton, 2019 ; Tosun & Senocak, 2013 ).

The Problem-Solving Method and Metacognition in Biology Instruction in Ethiopia

The National Education and Training Policy emphasizes the importance of education, particularly in science and technology, in improving problem-solving skills, cultural development, and environmental conservation for holistic development (ETP, 1994 ). Similarly, the 2009 Ethiopian Education Curriculum Framework Document highlights higher-order skills as key competencies and promotes the application, analysis, synthesis, evaluation, and innovation of knowledge for the twenty-first century (MoE, 2009 ). Whereas, a third revision of the curriculum is needed to promote science and technology studies with an emphasis on advanced cognitive skills and a shift from teacher-centered to learner-centered instructional methods (MoE, 2020 ).

The 2009 curriculum framework also places a strong emphasis on Biology as a life science, promoting understanding of self and living things while encouraging critical thinking and problem-solving. Biology lessons that integrate the problem-solving method can enhance students’ academic performance and understanding of the subject (Agaba, 2013 ). However, the Ethiopian education system faces challenges due to limited instructional resources, poor instructional methods, and a lack of experience in practical (hands-on) activities (Eshete, 2001; ETP, 1994 ; MoE, 2005 ; Negash, 2006 ). On the other hand, teachers’ inability to demonstrate effective instructional practices may contribute to low academic performance (Ganyaupfu, 2013 ; Umar, 2011 ).

Challenges in Implementing the PSMMS in Biology Instruction

Metacognitive processes are crucial for guiding learners in problem-solving activities (Sbhatu, 2006), but assessing them can be challenging due to their covert nature (Georghiades, 2000 ). Just like other areas of study, implementing metacognitive scaffolding of the problem-solving method in Biology instruction faces challenges such as complex learning, outdated skills, self-study, overloaded curricula, and limited resources, as shown in Table  1 .

Context Analysis in the Design of the PSMMS for Biology Instruction

Biology lessons are designed for different contexts and consider factors such as the learning environment, prior knowledge, background information, and cultural orientation (Reich et al., 2006 ). For this study, the three domains of context analysis (learners, learning, and learning task contexts) of Smith and Ragan’s (2005) instructional design model (as cited in Getenet, 2020 ) are adapted to design a context-based PSMMS method to generate authentic examples, strong scenarios, and instructional content, as shown in Table  2 .

Research Design

The study analyzed the learning context, including the available instructional resources and facilities in selected schools in Shambu Town, considering teachers’ and students’ perspectives using a mixed-methods research design (Creswell, 2009 ; Creswell & Creswell, 2018 ).

Study Participants

The study was conducted in public secondary schools in Shambu Town. Two schools, namely Shambu Secondary and Preparatory School (ShSPS) and Shambu Secondary School (ShSS), were selected using purposive sampling. Additionally, two Natural Sciences grade 11 sections, one from each school, were selected for instructional intervention based on feedback from context analysis to design an instructional approach, specifically the PSMMS in this study. Thus, all 12 Biology teachers and 80 eleventh-grade students participated in this study (see Table  4 ).

Data Collection Instruments and Procedure

To analyze the contexts to design a context-driven PSMMS for Biology instruction, data were collected using interviews, observations, and a questionnaire. Interviews were conducted to get insights from teachers, while observations were used to assess classroom instructions and instructional resources. Likewise, a questionnaire was administered to students to collect quantitative data on their opinions about the use of PSMMS in Biology instruction. The questionnaire, which was adapted from existing literature (Kallio et al., 2017 ; Rahmawati et al., 2018 ), was initially produced in English and subsequently translated into local language (Afan Oromo) with the help of both software (English to Oromo translator software) and experts. The questionnaire was pilot-tested on a sample of 40 students (22 males and 18 females) to identify any deficiencies in the measuring instrument, and responses were rated on a five-point Likert scale ranging from strongly agree ( N  = 5) to strongly disagree ( N  = 1). The reliability score of the questionnaire was determined to be 0.895, which is at a good level of acceptability.

In this design-based research (DBR) to design an instructional approach for context-driven PSMMS, the data collection process follows a context analysis procedure. Subsequently, the quantitative data collection method is based on the qualitative approach. Accordingly, assessing the context and literature was the first step in the research process. The qualitative approach used interviews and observations for data collection and was also used to identify instructional deficiencies and formulate questions for quantitative data collection.

Data Analysis

This context-based study used both qualitative and quantitative methods to analyze the data collected. In this context-based study, data analysis was conducted on the complex networks of contextual components (Wang & Hannafin, 2005 ). According to Table  2 , the domains of context analysis and key themes that emerged and were applied in this study are listed in Table  3 .

Qualitative data included interviews and notes recorded on the observation checklist. These were analyzed through thematic categorization. Each record was first transcribed, imported into Excel for filtering, and then sent back to Microsoft Word for highlighting. The transcripts were read several times to get a feel for the whole thing. The observation checklist was assessed by watching video recordings and taking notes. However, SPSS software version 24.0 was used to analyze quantitative data using descriptive and inferential statistics, including frequency, percentage, mean, standard deviation, and one-sample t-test.

Results and Discussions

In the study, a total of 12 Biology teachers participated, with 11 males and one female. As displayed in Table  4 , 41.67% of the teacher participants were from ShSPS, while 58.33% were from ShSS. The majority of these teachers had master’s degrees and had over ten years of teaching experience. As for the students involved, 52.5% were from ShSS and 47.5% were from ShSPS. The sex ratio among the students was 51.25% males and 48.75% females (Table  4 ).

Teachers’ Context Analysis

Beliefs about the practices of using the psmms in biology instruction.

The study analyzed teachers’ beliefs about the importance of the PSMMS in Biology instruction. Accordingly, most teachers interviewed (10 out of 12) stated that PSMMS improves students’ learning by enhancing their thinking skills, subject understanding, self-directed learning techniques, and behavior change, suggesting that it has a significant impact on students’ learning. About this, the study participant gave the following illustrative response:

In my opinion, using PSMMS in Biology classes improves students’ higher-order thinking skills by allowing them to understand and articulate problems in their context, stimulate reflection, and promote practical application knowledge (Teacher 4, ShSPS).

Concerning supportive learning, most of the teachers (nine out of 12) believed that it could enhance students’ engagement despite challenges in understanding and learning. About this, research participants said the following:

The PSMMS provides an engaging approach to Biology learning that promotes students’ active engagement and strengthens their awareness and understanding of the objectives and concepts they are expected to understand (Teacher 1, ShSS). Despite the challenge, I believe that using metacognitive scaffolding in the problem-solving method will help students develop their critical thinking skills. In addition, both teachers and students enjoy participating in the teaching-learning process in a classroom environment that is conducive to learning (Teacher 4, ShSPS).

The majority of teachers (eight out of 12) interviewed about PSMMS in Biology instruction argued that it is not commonly used in classrooms and instead relies on established methods like group discussions, pre-learning questions, projects, and quizzes. Some sample responses from teachers are:

The problem-solving method augmented by metacognition is crucial to learning Biology, although students and teachers have limited experience. However, motivated students using this strategy can make the Biology learning experience attractive (Teacher 2, ShSPS). Most students find learning Biology through the PSMMS a tiresome activity and believe that it is too challenging to achieve their learning goals (Teacher 1, ShSPS). The inability to implement the PSMMS in Biology learning experiences is attributed to inadequate laboratory equipment, teaching aids, and school facilities (Teacher 7, ShSS). On some occasions, I provide students with classwork, plans for implementing teaching strategies, arrange group discussions, and assist them in practicing subject-related skills. I then provide background information, promote class engagement, guide responses to questions, assess students’ existing knowledge and goals, provide relevant comments, and guide their thinking (Teacher 4, ShSPS).

Based on the results of the data analysis, it was found that teachers’ perceptions of the importance of the PSMMS to students’ Biology learning contributed significantly to the analysis of the learning context. Accordingly, the contribution of the PSMMS was to enhance students’ Biology learning by improving their critical thinking and learning experiences. Consistent with these findings, teachers’ positive beliefs about classroom problem-solving processes influence their approach to effective Biology teaching (Ishaku, 2015), and integrating metacognitive classroom interventions improves student learning, as evidenced by changes in conceptual learning and problem-solving skills (Guterman, 2002 ; Howard et al., 2001 ).

Observation of Teachers’ Classroom Instruction

The classroom instructional situation was observed to examine the effectiveness of PSMMS for Biology instruction. Consequently, teachers’ use of the PSMMS in Biology lessons was observed. According to the observation checklist, a total of 12 lessons, each lasting 40 minutes, were audited. The first step was to examine teachers’ daily lesson plans. Objectives were found to center predominantly on cognitive domains, neglecting higher-order problem-solving and metacognitive skills. This was evident from the use of terms such as “understand,” “know,” “write,” “explain,” and “describe” in the lesson plan objectives, which hold little significance for teaching Biology using the PSMMS. This finding is consistent with previous research (Chandio et al., 2016 ; Hyder & Bhamani, 2016 ) showing that the objectives of classroom lesson plans often focus on the lower cognitive domain, indicating lower-level knowledge acquisition.

Observing how teachers deliver lessons in the classroom revealed that they often require students to participate in group discussions, which they believe is a learner-centered approach. However, student engagement was limited, and the details of the tasks that students were expected to discuss were not outlined. Additionally, in the lessons observed, teachers failed to engage students, connect theory with practical applications, or support activity-based learning. On the other hand, teachers still have limited opportunities to assess understanding through targeted questions and encourage the use of critical thinking skills. Only oral questions, tests, or quizzes are used as an assessment method. These results were contradictory to the findings of other researchers’ studies, such as Ahmady and Nakhostin-Ruhi ( 2014 ) and Ishaku (2015), where teachers’ classroom lesson delivery is based on students’ constructivist and learner-centered environment acquiring advanced and critical thinking skills from Biology lessons.

The observation raised further questions regarding multimodal lesson delivery, revealing the use of visual representations of figures and diagrams in addition to the usual lecture style (auditory), raising additional concerns about multimodal instructional delivery. Therefore, there was no way to verify whether students had acquired the required higher-order skills, such as problem-solving and metacognitive skills, during their Biology learning. This finding contradicts the findings of Syofyan and Siwi’s ( 2018 ) research, which claims that students’ learning approaches are influenced by their sensory experiences. Consequently, students employ all their senses to capture information when teachers employ visual, auditory, and kinesthetic learning styles.

Students’ Context Analysis

The section presents the results of students’ responses collected using survey questions. Using a questionnaire with a five-point Likert scale ranging from strongly agree to strongly disagree (5 = strongly agree, 4 = agree, 3 = neutral, 2 = disagree, and 1 = strongly disagree), the impact of using PSMMS in Biology learning practices on students’ problem-solving and metacognitive skills was examined. The questionnaire had a response rate of 80 out of 98 (81.63%), indicating satisfactory status and acceptable use of the instrument. Therefore, in students’ responses to the survey questions on Biology learning practices using the PSMMS, there is significant ( p  < 0.05) variation across all dimensions of the items (M = 4.32, SD = 1.30), with mean scores above 4 indicating general students’ agreement with most items listed in Table  5 .

Regarding the problem-solving skills (Items 1–5) that students would acquire in their Biology learning practices using the PSMMS in Biology lessons, the strongest agreement was to investigate and identify the most effective problem-solving strategies (Item 4, M = 4.25, SD = 1.11), followed by creating the framework and design of the problem-solving activities (Item 2, M = 4.05, SD = 1.16), appropriately evaluating the results and providing alternative solutions to the problems (Item 5, M = 3.91, SD = 1.21), and identifying the problem in the problem sketch and interpreting the final result (Item 1, M = 3.90, SD = 1.28). On the other hand, students typically expressed less positive views about the PSMMS’s use of Biology instruction to enhance laboratory knowledge and problem-solving skills (Item 3, M = 3.25, SD = 1.57), despite significant differences in response patterns (Table  5 ).

Concerning students’ responses to the questionnaire items on metacognitive skills (Items 6–15) acquired in their Biology learning practices using the PSMMS, Table  5 shows that the most positive item states that the use of the PSMMS helps set clear learning objectives (Item 7, M = 4.36, SD = 1.09) and evaluates success by asking how well they did (Item 15, M = 4.29, SD = 1.10). Students tended to be less positive about learning Biology using the PSMMS, which is used to create examples and diagrams to make information more meaningful (Item 9, M = 3.83, SD = 1.21), despite the wide range of response patterns (Table  5 ). As a result, using PSMMS in Biology instruction helps students learn essential planning (Items 6–8), implementing (Items 9 and 10), monitoring (Items 11 and 12), and evaluating (Items 13–15) strategies for practice and to learn real-world applications of Biology (Table  5 ).

After data analysis of students’ responses to the survey questions, it was found that the PSMMS instructional approach is effective in helping students acquire problem-solving and metacognitive skills in their Biology learning practices. However, teachers’ responses, classroom observations, and resource availability indicated that the PSMMS approach was not effectively used to improve students’ problem-solving skills and strategies in Biology learning. The study highlights the disadvantages of shortages of laboratory facilities and large class sizes when implementing learner-centered practices in schools. These issues are supported by Kawishe’s (2016) study. Additionally, the PSMMS was not effectively applied in Biology instruction, resulting in students’ inability to develop metacognitive strategies and skills. Therefore, as studies have shown, students face challenges in acquiring metacognitive knowledge and regulation, which are crucial for the development of higher-order thinking skills in Biology learning (Aaltonen & Ikavalko, 2002 ; Lai, 2011 ).

Learning Context Analysis

This section presents the learning context analysis of PSMMS-based Biology instruction for two aspects, namely the availability of instructional resources in laboratories and pedagogical centers and the challenges in implementing the PSMMS in Biology instruction at Shambu Secondary and Preparatory School (ShSPS) and Shambu Secondary School (ShSS). Each is described below.

Availability of Instructional Resources in the Laboratories and Pedagogical Centers

In this section, a physical observation was conducted to assess the availability of instructional resources in Biology laboratories and pedagogical centers. The observation checklists were used to examine the impacts of their availability on Biology instruction using PSMMS.

Concerning the observations of the laboratory resources, it was noted that the two schools have independent Biology laboratories, but their functioning is hindered by poor organization, display tables, and a lack of water supply and waste disposal systems, as shown in Table  6 . Some basic laboratory equipment and chemicals, including dissecting kits, centrifuges, measuring cylinders, protein foods, sodium hydroxide solution, 1% copper (II) sulfate solution, gas syringes, and hydrogen peroxide, are missing. One school, ShSS, has only seven resources out of 20 identified for observation, making it difficult to conduct laboratory activities (Table  6 ).

Regarding the observations of instructional or teaching resources in the pedagogical centers, the results are shown in Table  7 . The results showed that there were no independent or autonomous pedagogical centers in the two schools; instead, they used the Biology department offices as a pedagogical center and kept some teaching and learning aids there. On the other hand, only DNA and RNA models were accessible in ShSPS, while models of DNA and RNA as well as illustrations depicting the organization of animal cell structures were available in ShSS (Table  7 ).

Challenges of Using the PSMMS in Biology Instruction

In this case, the results of interviews with teachers and survey results from students about the challenges they encountered when using the PSMMS in Biology instruction were used. The results of teachers’ and students’ responses are described below.

Teachers’ interview responses regarding the challenges they encountered in implementing the PSMMS in Biology instruction served as the basis for teachers’ perspectives . With the exception of two teachers who gave insignificant responses, the other teachers’ responses were categorized thematically. Therefore, Table  8 contains the response categories by themes, the number of respondents (N), and examples of responses. According to most teachers ( N  = 10), there is a lack of the required up-to-date knowledge, skills, and experience, and for other teachers ( N  = 7), there are shortages of equipment and chemicals (in Biology laboratories) as well as instructional aids (in pedagogical centers), which are challenges of using the PSMMS in Biology instruction. They also mentioned that challenging factors, such as the high student-teacher ratio and time constraints ( N  = 4), students’ deficiency of knowledge and attitudes towards learning ( N  = 3), and problems with school administrative functions ( N  = 1), have an impact on how well students learn Biology while using the PSMMS instructional approach (Table  8 ).

Students’ perspectives , however, were based on their responses to survey questions concerning the challenges of using the PSMMS in Biology lessons, as shown in Table  9 below. The study found statistically significant ( p  < 0.05) differences across the five-item dimensions, with an average mean of 3.62 and a standard deviation of 1.36. Consequently, mean scores above 3 indicated that students agreed with the challenges of implementing the PSMMS in Biology instruction (Table  9 ).

As shown in Table  9 , the majority of students identified two key challenges to successfully implementing the PSMMS in their learning. These are shortages of instructional resources (Item 2, M = 3.56, SD = 1.39) and student difficulty in connecting their prior knowledge with Biological concepts (Item 1, M = 3.44, SD = 1.42). On the other hand, students responded that their teachers had the knowledge and awareness to conduct instructional processes using the PSMMS (Item 4, M = 3.95, SD = 1.22) and had the skills and competence to conduct instructional processes using the PSMMS (Item 5, M = 3.98, SD = 1.35). Table  9 also shows that, despite significant differences in response patterns, students generally had a negative opinion about the dominance of some students in collaborative work (Item 3, M = 3.16, SD = 1.43).

According to the analyzed data, one of the challenging factors was that teachers often lack the required knowledge and skills to facilitate learning, scaffold it, and successfully implement PSMMS in Biology instruction. In contrast, Belland et al. ( 2013 ) suggested that instructional scaffolds increase students’ autonomy, competence, and intimacy, which improves their motivation and enables them to identify appropriate challenges. The other challenging factor that influenced the use of the PSMMS in Biology instruction was the shortage of instructional resources and facilities. Consistent with the studies of Daganaso et al. ( 2020 ) and Kawishe (2016), the use of the PSMMS for Biology instruction faces challenges due to inadequate instructional resources, time constraints, and large class sizes. However, as Eshete (2001) describes, students lack the importance of instructional resources, as instructional resources are necessary for students to learn Biology effectively as they are essential for a deeper understanding of science.

Generally, the important findings from the analyses of the teachers, learners, and learning contexts and their implications for design principles are summarized in Table  10 .

Conclusions

In this study, contexts (teachers, students, and learning) were analyzed with the aim of designing a context-driven problem-solving method with metacognitive scaffolding (PSMMS) for Biology instruction. Despite the potential benefits of the PSMMS, the findings of the current study indicate that the use of the PSMMS instructional approach faces challenges. These challenges include teachers’ lack of the required up-to-date knowledge and skills, students’ lack of awareness and positive attitude towards learning, an overloaded curriculum, scarcity of resources, large class sizes, and problems with school administrative functions. The study emphasizes the significance of context analysis in the design of an effective PSMMS instructional method for enhancing students’ learning in Biology. This analysis provides useful information for providing pertinent examples, practical content, and context-driven instruction.

The context-driven instructional design approach, using the PSMMS, addresses problems in teachers’ effectiveness, students’ effective learning, and the establishment of supportive teaching and learning environments. This approach considers the performance of both teachers and students, as well as the learning environment, including the availability of instructional resources. Consequently, this study concludes that understanding the needs of teachers in relation to the PSMMS can help both teachers and educational policymakers design a system that is well-suited to their specific requirements. Additionally, it can help students use their practical skills as well as establish connections between their prior knowledge and the Biology concepts they are learning. This process has the potential to generate innovative systems for applying the PSMMS instructional approach, with teachers serving as facilitators and students actively engaging and taking responsibility for their own learning progress.

The study investigated the importance of incorporating target groups into the design of the PSMMS for Biology instruction. The study’s empirical findings support the notion that the PSMMS should provide regular learning opportunities and foster the active engagement of teachers. The study also emphasizes the need to consider learning contexts while designing the PSMMS for Biology instruction that is deeply rooted in its particular context, as effective principles applied in one context could not yield the same results in another context. The study suggests that this strategy is particularly useful in developing countries like Ethiopia, where there is limited experience with metacognitive strategies to scaffold the problem-solving method in Biology instruction. As a result, the authors recommend expanding the target audience, considering the national context, and incorporating metacognitive knowledge and regulation strategies in designing context-driven PSMMS for secondary school Biology instruction.

Data Availability

The authors confirm that the results of this study are available in the article and its supplementary material, and raw data can be obtained from the corresponding author upon reasonable request.

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Acknowledgements

The authors would like to thank the teachers and students of Shambu Secondary Schools, Jimma University, and Shambu College of Teachers Education for their invaluable contributions in terms of information, resources, and financial support.

This editorial has not received financial support from any funding organizations.

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College of Natural Sciences, Department of Biology, Jimma University, Jimma, Ethiopia

Merga Dinssa Eticha & Tsige Ketema

Department of Biology, Shambu College of Teachers Education, Shambu, Ethiopia

Merga Dinssa Eticha

Department of Curriculum and Instructional Sciences, Kotebe University of Education, Addis Ababa, Ethiopia

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Contributions

Merga Dinssa Eticha : Conceptualization, Methodology, Validation, Formal analysis, Investigation, Resources, Writing-original draft, Writing-review and editing.

Adula Bekele Hunde : Conceptualization, Methodology, Validation, Investigation, Supervision, Writing-review and editing.

Tsige Ketema : Conceptualization, Methodology, Validation, Investigation, Supervision, Writing-review and editing.

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Correspondence to Merga Dinssa Eticha .

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Eticha, M.D., Hunde, A.B. & Ketema, T. Designing a Context-Driven Problem-Solving Method with Metacognitive Scaffolding Experience Intervention for Biology Instruction. J Sci Educ Technol (2024). https://doi.org/10.1007/s10956-024-10107-x

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Accepted : 27 February 2024

Published : 27 August 2024

DOI : https://doi.org/10.1007/s10956-024-10107-x

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