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Lesson 6: Assigning Probabilities to Events; Probability Rules

Hi everyone! Read through the material below, watch the videos, work on the Word lecture and follow up with your instructor if you have questions.

You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.

WeBWorK. Set 3.1

Learning Outcomes

  • Understand and use the terminology of probability.
  • Determine whether two events are mutually exclusive and whether two events are independent.
  • Calculate probabilities using the Addition Rules and Multiplication Rules.
  • Construct and interpret Venn Diagrams.
  • Construct and interpret Tree Diagrams.

Common Resoursces

  • 3.1 Terminology
  • 3.2 Independent and Mutually Exclusive Events
  • 3.3 Two Basic Rules of Probability
  • 3.5 Tree and Venn Diagram
  • Lesson06-ProbabilityBasics
  • WordLecture-06
  • Open Intro Slides

Basic Probability, Definitions, and Venn Diagrams

Properties of Probability, the Addition Rule

Excel-based Course Resources

  • Introductory Statistics by Sheldon Ross, 3rd edition: Sections 4.1-4.3

R-based Course Resources

  • R-laboratory

Watch the video introduction to probability .

  • What is a random phenomenon?
  • Explain why weather is an example of a random phenomenon.
  • What does it mean when a weather reporter says that there is a 70% chance of rain tomorrow?
  • If we flip a fair coin repeatedly, what can be said about the proportion of heads in the short run? What can be said about the proportion of heads in the long run?
  • What can you say about an event whose probability is close to one compared to an event whose probability is close to zero?

Consider three sets A , B , and C . What does $n(A\cup B \cup C)$ look like?

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Module 6: Probability and Probability Distributions

Introduction to Probability Rules

What you’ll learn to do: reason from probability distributions, using probability rules, to answer probability questions..

In this section, we introduce probability rules and properties. These rules can make evaluating probabilities far simpler and can also help catch mistakes if results are nonsensical (for example, a 140% chance is impossible). We revisit conditional probabilities, which are a fundamental concept in understanding how to interpret results from hypothesis testing. Finally, we introduce the notion of independence, joint, and marginal probabilities, and present a useful rule that ties these concepts together.

  • Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

Concepts in Statistics Copyright © 2023 by CUNY School of Professional Studies is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.

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2.4 - how to assign probability to events.

We know that probability is a number between 0 and 1. How does an event get assigned a particular probability value? Well, there are three ways of doing so:

  • the personal opinion approach
  • the relative frequency approach
  • the classical approach

On this page, we'll take a look at each approach.

The Personal Opinion Approach Section  

This approach is the simplest in practice, but therefore it also the least reliable. You might think of it as the "whatever it is to you" approach. Here are some examples:

  • "I think there is an 80% chance of rain today."
  • "I think there is a 50% chance that the world's oil reserves will be depleted by the year 2100."
  • "I think there is a 1% chance that the men's basketball team will end up in the Final Four sometime this decade."

Example 2-4 Section  

At which end of the probability scale would you put the probability that:

  • one day you will die?
  • you can swim around the world in 30 hours?
  • you will win the lottery someday?
  • a randomly selected student will get an A in this course?
  • you will get an A in this course?

The Relative Frequency Approach Section  

The relative frequency approach involves taking the follow three steps in order to determine P ( A ), the probability of an event A :

  • Perform an experiment a large number of times, n , say.
  • Count the number of times the event A of interest occurs, call the number N ( A ), say.
  • Then, the probability of event A equals:

\(P(A)=\dfrac{N(A)}{n}\)

The relative frequency approach is useful when the classical approach that is described next can't be used.

Example 2-5 Section  

Penny

When you toss a fair coin with one side designated as a "head" and the other side designated as a "tail", what is the probability of getting a head?

I think you all might instinctively reply \(\dfrac{1}{2}\). Of course, right? Well, there are three people who once felt compelled to determine the probability of getting a head using the relative frequency approach:

As you can see, the relative frequency approach yields a pretty good approximation to the 0.50 probability that we would all expect of a fair coin. Perhaps this example also illustrates the large number of times an experiment has to be conducted in order to get reliable results when using the relative frequency approach.

By the way, Count Buffon (1707-1788) was a French naturalist and mathematician who often pondered interesting probability problems. His most famous question

Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips?

came to be known as Buffon's needle problem. Karl Pearson (1857-1936) effectively established the field of mathematical statistics. And, once you hear John Kerrich's story, you might understand why he, of all people, carried out such a mind-numbing experiment. He was an English mathematician who was lecturing at the University of Copenhagen when World War II broke out. He was arrested by the Germans and spent the war interned in a prison camp in Denmark. To help pass the time he performed a number of probability experiments, such as this coin-tossing one.

Example 2-6 Section  

trees

Some trees in a forest were showing signs of disease. A random sample of 200 trees of various sizes was examined yielding the following results:

What is the probability that one tree selected at random is large?

There are 68 large trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is large is 68/200 = 0.34.

What is the probability that one tree selected at random is diseased?

There are 37 diseased trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is diseased is 37/200 = 0.185.

What is the probability that one tree selected at random is both small and diseased?

There are 8 small, diseased trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is small and diseased is 8/200 = 0.04.

What is the probability that one tree selected at random is either small or disease-free?

There are 121 trees (35 + 46 + 24 + 8 + 8) out of 200 total trees that are either small or disease-free, so the relative frequency approach would tell us that the probability that a tree selected at random is either small or disease-free is 121/200 = 0.605.

What is the probability that one tree selected at random from the population of medium trees is doubtful of disease?

There are 92 medium trees in the sample. Of those 92 medium trees, 32 have been identified as being doubtful of disease. Therefore, the relative frequency approach would tell us that the probability that a medium tree selected at random is doubtful of disease is 32/92 = 0.348.

The Classical Approach Section  

The classical approach is the method that we will investigate quite extensively in the next lesson. As long as the outcomes in the sample space are equally likely (!!!), the probability of event \(A\) is:

\(P(A)=\dfrac{N(A)}{N(\mathbf{S})}\)

where \(N(A)\) is the number of elements in the event \(A\), and \(N(\mathbf{S})\) is the number of elements in the sample space \(\mathbf{S}\). Let's take a look at an example.

Example 2-7 Section  

Suppose you draw one card at random from a standard deck of 52 cards. Recall that a standard deck of cards contains 13 face values (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King) in 4 different suits (Clubs, Diamonds, Hearts, and Spades) for a total of 52 cards. Assume the cards were manufactured to ensure that each outcome is equally likely with a probability of 1/52. Let \(A\) be the event that the card drawn is a 2, 3, or 7. Let \(B\) be the event that the card is a 2 of hearts (H), 3 of diamonds (D), 8 of spades (S) or king of clubs (C). That is:

  • \(A= \{x: x \text{ is a }2, 3,\text{ or }7\}\)
  • \(B = \{x: x\text{ is 2H, 3D, 8S, or KC}\}\)
  • What is the probability that a 2, 3, or 7 is drawn?
  • What is the probability that the card is a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
  • What is the probability that the card is either a 2, 3, or 7 or a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
  • What is \(P(A\cap B)\)?

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This page should be used as a quick reference guide for this unit. Some terminology and concepts presented here are covered in more detail on other pages. Refer back to these rules as needed.

1) Possible values for probabilities range from 0 to 1

0 = impossible event 1 = certain event

2) The sum of all the probabilities for all possible outcomes is equal to 1.

Note the connection to the complement rule.

3) Addition Rule - the probability that one or both events occur

mutually exclusive events: P(A or B) = P(A) + P(B) not mutually exclusive events: P(A or B) = P(A) + P(B) - P(A and B)

4) Multiplication Rule - the probability that both events occur together

independent events: P(A and B) = P(A) * P(B) P(A and B) = P(A) * P(B|A)

5) Conditional Probability - the probability of an event happening  given  that another event has already happened

P(A|B) = P(A and B) / P(B) *Note the line | means "given" while the slash / means divide

Key Terminology

Mutually Exclusive - this indicates that two events cannot happen at the same time.

For example, consider the following two events: A) rolling a 2 and B) rolling an odd number. Since 2 is an even number, it's not possible for me to roll a 2 and for that number to be odd. Therefore, these events are mutually exclusive.

Independent Events  - the probability of one event does not change based on the outcome of the other event

Consider a basketball player shooting 2 free throws. If the player's probability of making the second shot changes based on whether or not they make the first shot, then these events are dependent. If the probability does not change, then they would be independent.

  • << Previous: Probability
  • Next: Single Event Probability >>
  • Last Updated: Apr 19, 2024 3:09 PM
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7.10: Conditional Probability and the Multiplication Rule

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Two red dice are shown. The first die shows 5 and the second die shows 1.

Learning Objectives

After completing this section, you should be able to:

  • Calculate conditional probabilities.
  • Apply the Multiplication Rule for Probability to compute probabilities.

Back in Example 7.18, we constructed the following table (Figure 7.38) to help us find the probabilities associated with rolling two standard 6-sided dice:

A table with 6 rows and 6 columns. The columns represent the first die and are titled, 1, 2, 3, 4, 5, and 6. The rows represent the second die and are titled, 1, 2, 3, 4, 5, and 6. The data is as follows: Row 1: 2, 3, 4, 5, 6, 7. Row 2: 3, 4, 5, 6, 7, 8. Row 3: 4, 5, 6, 7, 8, 9. Row 4: 5, 6, 7, 8, 9, 10. Row 5: 6, 7, 8, 9, 10, 11. Row 6: 7, 8, 9, 10, 11, 12.

For example, 3 of these 36 equally likely outcomes correspond to rolling a sum of 10, so the probability of rolling a 10 is 3 36 = 1 12 3 36 = 1 12 . However, if you choose to roll the dice one at a time, the probability of rolling a 10 will change after the first die comes to rest. For example, if the first die shows a 5, then the probability of rolling a sum of 10 has jumped to 1 6 1 6 —the event will occur if the second die also shows a 5, which is 1 of 6 equally likely outcomes for the second die. If instead the first die shows a 3, then the probability of rolling a sum of 10 drops to 0—there are no outcomes for the second die that will give us a sum of 10.

Understanding how probabilities can shift as we learn new information is critical in the analysis of our second type of compound events: those built with “and.” This section will explain how to compute probabilities of those compound events.

Conditional Probabilities

When we analyze experiments with multiple stages, we often update the probabilities of the possible final outcomes or the later stages of the experiment based on the results of one or more of the initial stages. These updated probabilities are called conditional probabilities .

In other words, if O O is a possible outcome of the first stage in a multistage experiment, then the probability of an event E E conditional on O O (denoted P ( E | O ) P ( E | O ) , read “the probability of E E given O O ”) is the updated probability of E E under the assumption that O O occurred.

In the example that opened this section, we might consider rolling two dice as a multistage experiment: rolling one, then the other. If we define E E to be the event “roll a sum of 10,” O O to be the event “first die shows 5,” and Q Q to be the event “first die shows 3,” then we computed P ( E ) = 1 12 P ( E ) = 1 12 , P ( E | O ) = 1 6 P ( E | O ) = 1 6 , and P ( E | Q ) = 0 P ( E | Q ) = 0 .

Example 7.31

Computing conditional probabilities.

  • April is playing a coin-flipping game with Ben. She will flip a coin 3 times. If the coin lands on heads more than tails, April wins; if it lands on tails more than heads, Ben wins. Let A A be the event “April wins,” H H be “first flip is heads,” and T T be “first flip is tails.” Compute P ( A ) P ( A ) , P ( A | H ) P ( A | H ) , and P ( A | T ) P ( A | T ) .
  • You are about to draw 2 cards without replacement from a deck containing only these 10 cards: A ♡ A ♡ , A ♠ A ♠ , A ♣ A ♣ , A ♢ A ♢ , K ♠ K ♠ , K ♣ K ♣ , Q ♡ Q ♡ , Q ♠ Q ♠ , J ♡ J ♡ , J ♠ J ♠ . We’ll define the following events: F F is “both cards are the same rank,” A A is “first card is an ace,” and K K is “first card is a king.” Compute P ( F | A ) P ( F | A ) and P ( F | K ) P ( F | K ) .
  • Jim’s sock drawer contains 5 black socks and 3 blue socks. To avoid waking his partner, Jim doesn’t want to turn the lights on, so he puts on 2 socks at random. Let M M be the event “Jim’s 2 socks match,” let K K be the event “the sock on Jim’s left foot is black,” and let L L be the event “the sock on Jim’s left foot is blue.” Compute P ( M ) P ( M ) , P ( M | K ) P ( M | K ) , and P ( M | L ) P ( M | L ) .

Step 2. Now, let’s compute P ( A | H ) P ( A | H ) . We are assuming the result of the first flip is heads. That leaves us with 4 possible outcomes: HHH, HHT, HTH, and HTT. Of those, April wins 3 (HHH, HHT, HTH) and loses one (HTT). So, P ( A | H ) = 3 4 P ( A | H ) = 3 4 .

Step 3. If the result of the first flip is instead tails, the 4 possible outcomes are THH, THT, TTH, and TTT. Of those, April wins 1 (THH) and loses 3 (THT, TTH, TTT). So, P ( A | T ) = 1 4 P ( A | T ) = 1 4 .

Step 2. If the event K K happens instead, then the first card drawn is a king. That leaves 4 aces, 1 king, 2 queens, and 2 jacks in the deck. Under the assumption that the first card is a king, the event F F will occur only if the second card is also a king. Since only one of the remaining 9 cards is a king, we have P ( F | K ) = 1 9 P ( F | K ) = 1 9 .

Step 2. If the sock on Jim’s left foot is black (i.e., K K occurred), then there are 4 remaining black socks of the 7 in the drawer. So, P ( M | K ) = 4 7 P ( M | K ) = 4 7 .

Step 3. If the sock on Jim’s left foot is blue ( L L occurred), then there are 2 blue socks among the 7 remaining in the drawer. So, P ( M | L ) = 2 7 P ( M | L ) = 2 7 .

Your Turn 7.31

In Tree Diagrams, Tables, and Outcomes, we introduced the concept of dependence between stages of a multistage experiment. We stated at the time that two stages were dependent if the result of one stage affects the other stage. We explained that dependence in terms of the sample space, but sometimes that dependence can be a little more subtle; it’s more properly understood in terms of conditional probabilities. Two stages of an experiment are dependent if P ( E | F ) ≠ P ( E | F ′ ) P ( E | F ) ≠ P ( E | F ′ ) for some outcome of the second stage E E and outcome of the first stage F F .

Protecting Bombers in World War II

In his book How Not to Be Wrong , Jordan Ellenberg recounts this anecdote: During World War II, the American military wanted to add additional armor plating to bomber aircraft, in order to reduce the chances that they get shot down. So, they collected data on planes after returning from missions. The data showed that the fuselage, wings, and fuel system had many more bullet holes (per unit area) than the engine compartments, so the military brass wanted to add additional armor to the parts of the plane that were hit most often. Luckily, before they added the armor to the planes, they asked for a second opinion. Abraham Wald, a Jewish mathematician who had fled the rising Nazi regime, pointed out that it was far more important that the armor plating be added to areas where there were fewer bullet holes. Why? The planes they were studying had already completed their missions, so the military was essentially looking at conditional probabilities: the probability of suffering a bullet strike, given that the plane made it back safely. More bullet holes in an area on the plane indicated that was a region that wasn’t as important for the plane’s survival!

Compound Events Using “And” and the Multiplication Rule

For multistage experiments, the outcomes of the experiment as a whole are often stated in terms of the outcomes of the individual stages. Commonly, those statements are joined with “and.” For example, in the sock drawer example just above, one outcome might be “the left sock is black and the right sock is blue.” As with “or” compound events, these probabilities can be computed with basic arithmetic.

Multiplication Rule for Probability: If E E and F F are events associated with the first and second stages of an experiment, then P ( E and ⁢ F ) = P ( E ) × P ( F | E ) P ( E and ⁢ F ) = P ( E ) × P ( F | E ) .

In The Addition Rule for Probability, we considered probabilities of events connected with “and” in the statement of the Inclusion/Exclusion Principle. These two scenarios are different; in the statement of the Inclusion/Exclusion Principle, the events connected with “and” are both events associated with the same single-stage experiment (or the same stage of a multistage experiment). In the Multiplication Rule, we’re looking at events associated with different stages of a multistage experiment.

Example 7.32

Using the multiplication rule for probability.

You are president of a club with 10 members: 4 seniors, 3 juniors, 2 sophomores, and 1 first-year. You need to choose 2 members to represent the club on 2 college committees. The first person selected will be on the Club Awards Committee and the second will be on the New Club Orientation Committee. The same person cannot be selected for both. You decide to select these representatives at random.

  • What is the probability that a senior is chosen for both positions?
  • What is the probability that a junior is chosen first and a sophomore is chosen second?
  • What is the probability that a sophomore is chosen first and a senior is chosen second?
  • We need the probability that a senior is chosen first and a senior is chosen second. These are two stages of a multistage experiment, so we’ll apply the Multiplication Rule for Probability: P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . P ( senior chosen first and senior chosen second ) = P ( senior chosen first ) × P ( senior chosen second | senior chosen first ) . Since there are 4 seniors among the 10 members, P ( senior chosen first ) = 4 10 = 2 5 P ( senior chosen first ) = 4 10 = 2 5 . Next, assuming a senior is chosen first, there are 3 seniors among the 9 remaining members. So, P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 P ( senior chosen second | senior chosen first ) = 3 9 = 1 3 . Putting this all together, we get P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 P ( senior chosen first and senior chosen second ) = 2 5 × 1 3 = 2 15 .
  • There are 3 juniors among the 10 members, so P ( junior chosen first ) = 3 10 P ( junior chosen first ) = 3 10 . Assuming a junior is chosen first, there are 2 sophomores among the remaining 9 members, so P ( sophomore chosen second | junior chosen first ) = 2 9 P ( sophomore chosen second | junior chosen first ) = 2 9 . Thus, using the Multiplication Rule for Probability, we have P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 P ( junior chosen first and sophomore chosen second ) = 3 10 × 2 9 = 1 15 .
  • The probability that a sophomore is chosen first is 2 10 = 1 5 2 10 = 1 5 , and the probability that a senior is chosen second given that a sophomore was chosen first is 4 9 4 9 . Thus, using the Multiplication Rule for Probability, we have: P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 P ( sophomore chosen first and senior chosen second ) = 1 5 × 4 9 = 4 45 .

Your Turn 7.32

Work it out, the birthday problem.

One of the most famous problems in probability theory is the Birthday Problem, which has to do with shared birthdays in a large group. To make the analysis easier, we’ll ignore leap days, and assume that the probability of being born on any given date is 1 365 1 365 . Now, if you have 366 people in a room, we’re guaranteed to have at least one pair of people who share a single birthday. Imagine filling the room by first admitting someone born on January 1, then someone born on January 2, and so on… The 365th person admitted would be born on December 31. If you add one more person to the room, that person’s birthday would have to match someone else’s.

Let’s look at the other end of the spectrum. If you choose two people at random, what is the probability that they share a birthday? As with many probability questions, this is best addressed by find out the probability that they do not share a birthday. The first person’s birthday can be anything (probability 1), and the second person’s birthday can be anything other than the first person’s birthday (probability 364 365 364 365 ). The probability that they have different birthdays is 1 × 364 365 = 364 365 1 × 364 365 = 364 365 . So, the probability that they share a birthday is 1 − 364 365 = 1 365 1 − 364 365 = 1 365 .

What if we have three people? The probability that they all have different birthdays can be obtained by extending our previous calculation: The probability that two people have different birthdays is 364 365 364 365 , so if we add a third to the mix, the probability that they have a different birthday from the other two is 363 365 363 365 . So, the probability that all three have different birthdays is 364 365 × 363 365 ≈ 0.9918 364 365 × 363 365 ≈ 0.9918 , and thus the probability that there’s a shared birthday in the group is 1 − 0.9918 ≈ 0.0082 1 − 0.9918 ≈ 0.0082 .

The big question is this: How many people do we need in the room to have the probability of a shared birthday greater than 1 2 1 2 ? Make a guess, then with a partner keep adding hypothetical people to the group and computing probabilities until you get there!

It is often useful to combine the rules we’ve seen so far with the techniques we used for finding sample spaces. In particular, trees can be helpful when we want to identify the probabilities of every possible outcome in a multistage experiment. The next example will illustrate this.

Example 7.33

Using tree diagrams to help find probabilities.

The board game Clue uses a deck of 21 cards: 6 suspects, 6 weapons, and 9 rooms. Suppose you are about to draw 2 cards from this deck. There are 6 possible outcomes for the draw: 2 suspects, 2 weapons, 2 rooms, 1 suspect and 1 weapon, 1 suspect and 1 room, or 1 weapon and 1 room. What are the probabilities for each of these outcomes?

Step 1: Let’s start by building a tree diagram that illustrates both stages of this experiment. Let’s use S, W, and R to indicate drawing a suspect, weapon, and room, respectively (Figure 7.39).

A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage. The second stage represents the first card. The third stage representing the second card is as follows. The node, S branches into three nodes labeled S, W, and R. The node, W branches into three nodes labeled S, W, and R. The node, R branches into three nodes labeled S, W, and R. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.

Step 2: We want to start computing probabilities, starting with the first stage. The probability that the first card is a suspect is 6 21 = 2 7 6 21 = 2 7 . The probability that the first card is a weapon is the same: 2 7 2 7 . Finally, the probability that the first card is a room is 9 21 = 3 7 9 21 = 3 7 .

Step 3: Let’s incorporate those probabilities into our tree: label the edges going into each of the nodes representing the first-stage outcomes with the corresponding probabilities (Figure 7.40).

A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. The node, S branches into three nodes labeled S, W, and R. The node, W branches into three nodes labeled S, W, and R. The node, R branches into three nodes labeled S, W, and R. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.

Note that the sum of the probabilities coming out of the initial node is 1; this should always be the case for the probabilities coming out of any node!

Step 4: Let’s look at the case where the first card is a suspect. There are 3 edges emanating from that node (leading to the outcomes SS, SW, and SR). We’ll label those edges with the appropriate conditional probabilities, under the assumption that the first card is a suspect. First, there are 5 remaining suspect cards among the 20 left in the deck, so P ( second is suspect | first is suspect ) = 5 20 = 1 4 P ( second is suspect | first is suspect ) = 5 20 = 1 4 . Using similar reasoning, we can compute P ( second is weapon | first is suspect ) = 6 20 = 3 10 P ( second is weapon | first is suspect ) = 6 20 = 3 10 and P ( second is room | first is suspect ) = 9 20 P ( second is room | first is suspect ) = 9 20 .

Step 5: Checking our work, we see that the sum of these 3 probabilities is again equal to 1. Let’s add those to our tree (Figure 7.41).

A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R. The node, R branches into three nodes labeled S, W, and R. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.

Step 6: Let’s continue filling in the conditional probabilities at the other nodes, always checking to make sure the sum of the probabilities coming out of any node is equal to 1 (Figure 7.42).

A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R with the probabilities, three-tenths, one-fourth, and nine-twentieths. The node, R branches into three nodes labeled S, W, and R with the probabilities, three-tenths, three-tenths, and two-fifths. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R.

Step 7: We can compute the probability of landing on any final node by multiplying the probabilities along the path we would take to get there. For example, the probability of drawing a suspect first and a weapon second (i.e., ending up on the node labeled “SW”) is 2 7 × 3 10 = 3 35 Figure 7.43.

A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities, two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R with the probabilities three-tenths, one-fourth, and nine-twentieths. The node, R branches into three nodes labeled S, W, and R with the probabilities three-tenths, three-tenths, and two-fifths. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R. The path from the start node through the nodes, S and W is highlighted. The probability for the outcome, S W reads, P of S W equals 2 over 7 times 3 over 10 equals 3 over 35.

Step 8: Let’s fill in the rest of the probabilities (Figure 7.44).

A tree diagram with three stages. The diagram shows a node in the first stage branching into three nodes labeled S, W, and R in the second stage with the probabilities two-sevenths, two-sevenths, and three-sevenths, respectively. The second stage represents the first card. The third stage representing the second card is as follows. Node, S branches into three nodes labeled S, W, and R with the probabilities, one-fourth, three-tenths, and nine-twentieths. Node, W branches into three nodes labeled S, W, and R with the probabilities three-tenths, one-fourth, and nine-twentieths. The node, R branches into three nodes labeled S, W, and R with the probabilities three-tenths, three-tenths, and two-fifths. The possible outcomes are as follows: S S, S W, S R, W S, W W, W R, R S, R W, and R R. The probabilities for the outcomes are as follows. P of S S equals 2 over 7 times 1 over 4 equals 1 over 14. P of S W equals 2 over 7 times 3 over 10 equals 3 over 35. P of S R equals 2 over 7 times 9 over 20 equals 9 over 70. P of W S equals 2 over 7 times 3 over 10 equals 3 over 35. P of W W equals 2 over 7 times 1 over 4 equals 1 over 14. P of W R equals 2 over 7 times 9 over 20 equals 9 over 70. P of R S equals 3 over 7 times 3 over 10 equals 9 over 70. P of R W equals 3 over 7 times 3 over 10 equals 9 over 70. P of R R equals 3 over 7 times 2 over 5 equals 6 over 35.

Step 9: A helpful feature of tree diagrams is that the final outcomes are always mutually exclusive, so the Addition Rule can be directly applied. For example, the probability of drawing one suspect and one room (in any order) would be P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 P ( S R ) + P ( R S ) = 9 70 + 9 70 = 9 35 . We can find the probabilities of the other outcomes in a similar fashion, as shown in the following table:

Checking once again, the sum of these 6 probabilities is 1, as expected.

Your Turn 7.33

The monty hall problem.

On the original version of the game show Let’s Make a Deal , originally hosted by Monty Hall and now hosted by Wayne Brady, one contestant was chosen to play a game for the grand prize of the day (often a car). Here’s how it worked: On the stage were three areas concealed by numbered curtains. The car was hidden behind one of the curtains; the other two curtains hid worthless prizes (called “Zonks” on the show). The contestant would guess which curtain concealed the car. To build tension, Monty would then reveal what was behind one of the other curtains, which was always one of the Zonks (Since Monty knew where the car was hidden, he always had at least one Zonk curtain that hadn’t been chosen that he could reveal). Monty then turned to the contestant and asked: “Do you want to stick with your original choice, or do you want to switch your choice to the other curtain?” What should the contestant do? Does it matter?

With a partner or in a small group, simulate this game. You can do that with a small candy (the prize) hidden under one of three cups, or with three playing cards (just decide ahead of time which card represents the “Grand Prize”). One person plays the host, who knows where the prize is hidden. Another person plays the contestant and tries to guess where the prize is hidden. After the guess is made, the host should reveal a losing option that wasn’t chosen by the contestant. The contestant then has the option to stick with the original choice or switch to the other, unrevealed option. Play about 20 rounds, taking turns in each role and making sure that both contestant strategies (stick or switch) are used equally often. After each round, make a note of whether the contestant chose “stick” or “switch” and whether the contestant won or lost. Find the empirical probability of winning under each strategy. Then, see if you can use tree diagrams to verify your findings.

Check Your Understanding

Section 7.9 exercises.

  • /**/P\left({\text{tile shows A}}\right)/**/
  • /**/P\left(\text{tile shows A}\,|\,\text{tile shows a vowel}\right)/**/
  • /**/P({\text{tile shows a vowel}})/**/
  • /**/P\left({\text{tile shows a vowel}}\,|\,{\text{tile shows a letter that comes after M alphabetically}}\right)/**/

In the following exercises deal with the game “Punch a Bunch,” which appears on the TV game show The Price Is Right . In this game, contestants have a chance to punch through up to 4 paper circles on a board; behind each circle is a card with a dollar amount printed on it. There are 50 of these circles; the dollar amounts are given in this table:

Contestants are shown their selected dollar amounts one at a time, in the order selected. After each is revealed, the contestant is given the option of taking that amount of money or throwing it away in favor of the next amount. (You can watch the game being played in the video Playing “Punch a Bunch.” ) Jeremy is playing “Punch a Bunch” and gets 2 punches.

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4.5: Probability And Counting Rules

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  • Rupinder Sekhon and Roberta Bloom
  • De Anza College

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Learning Objectives

In this section, you will learn to:

  • Use probability tree diagrams to calculate probabilities
  • Use combinations to calculate probabilities

In this section, we will apply previously learnt counting techniques in calculating probabilities, and use tree diagrams to help us gain a better understanding of what is involved.

USING TREE DIAGRAMS TO CALCULATE PROBABILITIES

We already used tree diagrams to list events in a sample space. Tree diagrams can be helpful in organizing information in probability problems; they help provide a structure for understanding probability. In this section we expand our previous use of tree diagrams to situations in which the events in the sample space are not all equally likely.

We assign the appropriate probabilities to the events shown on the branches of the tree. By multiplying probabilities along a path through the tree, we can find probabilities for “and” events, which are intersections of events.

We begin with an example.

Example \(\PageIndex{1}\)

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn with replacement, what is the probability that both marbles are red?

Let \(\mathrm{E}\) be the event that the first marble drawn is red, and let \(\mathrm{F}\) be the event that the second marble drawn is red.

We need to find \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\).

By the statement, "two marbles are drawn with replacement," we mean that the first marble is replaced before the second marble is drawn.

There are 7 choices for the first draw. And since the first marble is replaced before the second is drawn, there are, again, seven choices for the second draw. Using the multiplication axiom, we conclude that the sample space \(\mathrm{S}\) consists of 49 ordered pairs. Of the 49 ordered pairs, there are \(3 \times 3 = 9\) ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

\[P(E \cap F)=\frac{9}{49} \nonumber \]

Further note that in this particular case

\[P(E \cap F)=\frac{9}{49}=\frac{3}{7} \cdot \frac{3}{7} \nonumber \]

giving us the result that in this example: \(\mathbf{P}(\mathbf{E} \cap \mathbf{F})=\mathbf{P}(\mathbf{E}) \cdot \mathbf{P}(\mathbf{F})\)

Example \(\PageIndex{2}\)

If in Example \(\PageIndex{1}\), the two marbles are drawn without replacement, then what is the probability that both marbles are red?

By the statement, "two marbles are drawn without replacement," we mean that the first marble is not replaced before the second marble is drawn.

Again, we need to find \(\mathrm{P}(\mathrm{E} \cap \mathrm{F})\).

There are, again, 7 choices for the first draw. And since the first marble is not replaced before the second is drawn, there are only six choices for the second draw. Using the multiplication axiom, we conclude that the sample space S consists of 42 ordered pairs. Of the 42 ordered pairs, there are \(3 \times 2 = 6\) ordered pairs that show red on the first draw and red on the second draw. Therefore,

\[P(E \cap F)=\frac{6}{42} \nonumber \]

Note that we can break this calculation down as

\[P(E \cap F)=\frac{6}{42}=\frac{3}{7} \cdot \frac{2}{6} \nonumber \].

Here 3/7 represents \(\mathrm{P}(\mathrm{E})\), and 2/6 represents the probability of drawing a red on the second draw, given that the first draw resulted in a red.

We write the latter as \(\mathrm{P}\)(red on the second | red on first) or \(\mathrm{P}(\mathrm{F} | \mathrm{E})\). The "|" represents the word "given" or “if”. This leads to the result that:

\[\mathbf{P}(\mathbf{E} \cap \mathbf{F})=\mathbf{P}(\mathbf{E}) \cdot \mathbf{P}(\mathbf{F} | \mathbf{E}) \nonumber \]

The is an important result, called the Multiplication Rule, which will appear again in later sections.

We now demonstrate the above results with a tree diagram.

Example \(\PageIndex{3}\)

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn without replacement, find the following probabilities using a tree diagram.

  • The probability that both marbles are red.
  • The probability that the first marble is red and the second white.
  • The probability that one marble is red and the other white.

Let \(\mathrm{R}\) be the event that the marble drawn is red, and let W be the event that the marble drawn is white.

We draw the following tree diagram.

Example8.3.3.png

  • The probability that both marbles are red is \(\mathrm{P}(\mathrm{RR})=6/42\)
  • The probability that the first marble is red and the second is white is \(\mathrm{P}(\mathrm{RW})=12/42\)
  • For the probability that one marble is red and the other is white, we observe that this can be satisfied if the first is red and the second is white, or if the first is white and the second is red. The “or” tells us we’ll be using the Addition Rule from Section 7.2.

Furthermore events \(\mathrm{RW}\) and \(\mathrm{WR}\) are mutually exclusive events, so we use the form of the Addition Rule that applies to mutually exclusive events.

\(\mathrm{P}\)(one marble is red and the other marble is white)

\[\begin{array}{l} =\mathrm{P}(\mathrm{RW} \text { or } \mathrm{WR}) \\ =\mathrm{P}(\mathrm{RW})+\mathrm{P}(\mathrm{WR}) \\ =12 / 42+12 / 42=24 / 42 \end{array} \nonumber \]

USING COMBINATIONS TO FIND PROBABILITIES

Although the tree diagrams give us better insight into a problem, they are not practical for problems where more than two or three things are chosen. In such cases, we use the concept of combinations that we learned in the last chapter. This method is best suited for problems where the order in which the objects are chosen is not important, and the objects are chosen without replacement.

Example \(\PageIndex{4}\)

Suppose a jar contains 3 red, 2 white, and 3 blue marbles. If three marbles are drawn without replacement, find the following probabilities.

  • \(\mathrm{P}\)(Two red and one white)
  • \(\mathrm{P}\)(One of each color)
  • \(\mathrm{P}\)(None blue)
  • \(\mathrm{P}\)(At least one blue)

Let us suppose the marbles are labeled as \(R_1,R_2,R_3,W_1,W_2,B_1,B_2,B_3\).

a. \(\mathrm{P}\)(Two red and one white)

Since we are choosing 3 marbles from a total of 8, there are 8\(\mathrm{C}\)3 = 56 possible combinations. Of these 56 combinations, there are \(3 \mathrm{C} 2 \times 2 \mathrm{C}1=6\) combinations consisting of 2 red and one white. Therefore,

\[P(\text { Two red and one white })=\frac{3 \mathrm{C} 2 \times 2 \mathrm{C} 1}{8 \mathrm{C} 3}=\frac{6}{56} \nonumber. \nonumber \]

b. \(\mathrm{P}\)(One of each color)

Again, there are 8\(\mathrm{C}\)3 = 56 possible combinations. Of these 56 combinations, there are \(3 \mathrm{Cl} \times 2 \mathrm{Cl} \times 3 \mathrm{Cl}=18\) combinations consisting of one red, one white, and one blue. Therefore,

\[P(\text { One of each color })=\frac{3 \mathrm{C} 1 \times 2 \mathrm{C} 1 \times 3 \mathrm{C} 1}{8 \mathrm{C} 3}=\frac{18}{56} \nonumber \]

c. \(\mathrm{P}\)(None blue)

There are 5 non-blue marbles, therefore

\[\mathrm{P}(\text { None blue })=\frac{5 \mathrm{C} 3}{8 \mathrm{C} 3}=\frac{10}{56}=\frac{5}{28} \nonumber \]

d. \(\mathrm{P}\)(At least one blue)

By "at least one blue marble," we mean the following: one blue marble and two non-blue marbles, OR two blue marbles and one non-blue marble, OR all three blue marbles. So we have to find the sum of the probabilities of all three cases.

\[\mathrm{P}(\mathrm{At} \text { least one blue })=\mathrm{P}(1 \text { blue, } 2 \text { non-blue) }+\mathrm{P}(2 \text { blue, l non-blue) }+\mathrm{P}(3\text { blue) } \nonumber \]

\[P( \text { At least one blue })=\frac{3 \mathrm{C} 1 \times 5 \mathrm{C} 2}{8 \mathrm{C} 3}+\frac{3 \mathrm{C} 2 \times 5 \mathrm{C} 1}{8 \mathrm{C} 3}+\frac{3 \mathrm{C} 3}{8 \mathrm{C} 3} \nonumber \]

\[ P(\text { At least one blue })=30 / 56+15 / 56+1 / 56=46 / 56=23 / 28 \nonumber \]

Alternately, we can use the fact that \(\mathrm{P}(\mathrm{E}) = 1 - \mathrm{P}(\mathrm{E}^c)\). If the event \(\mathrm{E}\) = At least one blue, then \(E^c\) = None blue.

But from part c of this example, we have \((\mathrm{E}^c) = 5/28\), so \(\mathrm{P}(\mathrm{E}) = 1 - 5/28 = 23/28\).

Example \(\PageIndex{5}\)

Five cards are drawn from a deck. Find the probability of obtaining two pairs, that is, two cards of one value, two of another value, and one other card.

Let us first do an easier problem-the probability of obtaining a pair of kings and queens.

Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is

\[\mathrm{P}(\mathrm{A} \text { pair of kings and queens })=\frac{4 \mathrm{C} 2 \times 4 \mathrm{C} 2 \times 44 \mathrm{C}1}{52 \mathrm{C} 5} \nonumber \]

To find the probability of obtaining two pairs, we have to consider all possible pairs.

Since there are altogether 13 values, that is, aces, deuces, and so on, there are 13\(\mathrm{C}\)2 different combinations of pairs.

\[P(\text { Two pairs })=13 \mathrm{C} 2 \cdot \frac{4 \mathrm{C} 2 \times 4 \mathrm{C} 2 \times 44 \mathrm{C}1}{52 \mathrm{C} 5}=.04754 \nonumber \]

Example \(\PageIndex{6}\)

A cell phone store receives a shipment of 15 cell phones that contains 8 iPhones and 7 Android phones. Suppose that 6 cell phones are randomly selected from this shipment. Find the probability that a randomly selected set of 6 cell phones consists of 2 iPhones and 4 Android phones.

There are 8\(\mathrm{C}\)2 ways of selecting 2 out of the 8 iPhones.

and 7\(\mathrm{C}\)4 ways of selecting 4 out of the 7 Android phones

But altogether there are 15\(\mathrm{C}\)6 ways of selecting 6 out of 15 cell phones.

Therefore we have

\[P(2 \text { iPhones and } 4 \text { Android phones })=\frac{8 \mathrm{C} 2 \times 7 \mathrm{C} 4}{15 \mathrm{C} 6}=\frac{(28)(35)}{5005}=\frac{980}{5005}=0.1958 \nonumber \]

Example \(\PageIndex{7}\)

One afternoon, a bagel store still has 53 bagels remaining: 20 plain, 15 poppyseed, and 18 sesame seed bagels. Suppose that the store owner packages up a bag of 9 bagels to bring home for tomorrow’s breakfast, and selects the bagels randomly. Find the probability that the bag contains 4 plain, 3 poppyseed, and 2 sesame seed.

There are 20\(\mathrm{C}\)4 ways of selecting 4 out of the 20 plain bagels,

and 15\(\mathrm{C}\)3 ways of selecting 3 out of the 15 poppyseed bagels,

and 18\(\mathrm{C}\)2 ways of selecting 2 out of the 18 sesame seed bagels.

But altogether there are 53\(\mathrm{C}\)9 ways of selecting 9 out of the 53 bagels.

\begin{array}{l} \mathrm{P} \text{(4 plain, 3 poppyseed, and 2 sesame seed)} &=\frac{20 \mathrm{C} 4 \times 15 \mathrm{C} 3 \times 18 \mathrm{C} 2}{53 \mathrm{C} 9} \\ &=\frac{(4845)(455)(153)}{4431613550} \\ &=0.761 \end{array}

We end the section by solving a famous problem called the Birthday Problem .

Example \(\PageIndex{8}\): Birthday Problem

If there are 25 people in a room, what is the probability that at least two people have the same birthday?

Let event \(\mathrm{E}\) represent that at least two people have the same birthday.

We first find the probability that no two people have the same birthday.

We analyze as follows.

Suppose there are 365 days to every year. According to the multiplication axiom, there are 365 25 possible birthdays for 25 people. Therefore, the sample space has 365 25 elements. We are interested in the probability that no two people have the same birthday. There are 365 possible choices for the first person and since the second person must have a different birthday, there are 364 choices for the second, 363 for the third, and so on. Therefore,

\[\mathrm{P}(\mathrm{No} \text { two have the same birthday })=\frac{365 \cdot 364 \cdot 363 \cdots 341}{365^{25}}=\frac{365 \mathrm{P} 25}{365^{25}} \nonumber \]

Since \(\mathrm{P}\)(at least two people have the same birthday) = 1 - \(\mathrm{P}\)(No two have the same birthday),

\[\mathrm{P} \text { at least two people have the same birthday ) }=1-\frac{365 \mathrm{P} 25}{365^{25}}=.5687\ \nonumber \]

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    Sum of all the probabilities should be 1. Construct a probability distribution. By using the knowledge that you have gained in section 4.2 and 4.5, list all possible values and associated probabilities. Find a probability of an event related with the X. Add the probabilities of the values of X that makes up the event.

  8. 7.1

    The opposite of "at least 3" is "getting a 1" (i.e. the only other possibility) so you can also figure the answer as 100% - 10% = 90% or 0.90. This rule of the opposites is our third rule of probability. Rule 3: The chance of something is 1 minus the chance of the opposite thing. Suppose you toss an astralgus twice.

  9. 3.3: Two Basic Rules of Probability

    This page titled 3.3: Two Basic Rules of Probability is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

  10. Lesson 6: Assigning Probabilities to Events; Probability Rules

    In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach. WeBWorK. Set 3.1.

  11. Introduction to Probability Rules

    In this section, we introduce probability rules and properties. These rules can make evaluating probabilities far simpler and can also help catch mistakes if results are nonsensical (for example, a 140% chance is impossible). We revisit conditional probabilities, which are a fundamental concept in understanding how to interpret results from ...

  12. 2.4

    The relative frequency approach involves taking the follow three steps in order to determine P ( A ), the probability of an event A: Perform an experiment a large number of times, n, say. Count the number of times the event A of interest occurs, call the number N ( A ), say. Then, the probability of event A equals: P ( A) = N ( A) n.

  13. PDF Probability Probability Rules

    We have two more formulae for probability that will be useful. Basic Probability Rules p(E) + p(E0) = 1, p(E [F) = p(E) + p(F) p(E \F) Use of Venn Diagrams for Probability. It is often helpful to put the information given in a Venn Diagram to organize the information and answer questions. The following is an example of such a case. Example 3.

  14. PDF 1. Assigning Probabilities

    1. Assigning Probabilities. The key to assigning probabilities is knowing all of your possible outcomes and knowing two rules: • All possible outcomes must total 1 or 100% (Where have we talked about 100% being important) • A probability must take a value 0 ≤ P(A) ≤ 1 (or 0% to 100%) 1) Probabilities can arise from empirical data, for ...

  15. LibGuides: Statistics Resources: Basic Probability Rules

    Refer back to these rules as needed. Basic Probability Rules. 1) Possible values for probabilities range from 0 to 1. 0 = impossible event. 1 = certain event. 2) The sum of all the probabilities for all possible outcomes is equal to 1. Note the connection to the complement rule. 3) Addition Rule - the probability that one or both events occur.

  16. PDF Lecture Notes 1 Basic Probability

    • Total Probability and Bayes Rule • Independence • Counting EE 178/278A: Basic Probability Page 1-1 ... • Probability law (measure or function) is an assignment of probabilities to events (subsets of sample space Ω) such that the following three axioms are satisfied: 1. P(A) ≥ 0, for all A(nonnegativity) ...

  17. 7.10: Conditional Probability and the Multiplication Rule

    Step 7: We can compute the probability of landing on any final node by multiplying the probabilities along the path we would take to get there. For example, the probability of drawing a suspect first and a weapon second (i.e., ending up on the node labeled "SW") is Unexpected text node: 'Figure 7.43.'. Figure 7.43.

  18. Flashcards Applying Probability Rules Assignment

    Applying Probability Rules Assignment. Log in. Sign up. A student surveyed 100 students and determined the number of students who take statistics or calculus among seniors and juniors. Here are the results.

  19. 4.3: The Addition and Multiplication Rules of Probability

    The multiplication rule: P(A AND B) = P(A|B)P(B) The addition rule: P(A OR B) = P(A) + P(B) − P(A AND B) Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder.

  20. 4.5: Probability And Counting Rules

    P(A pair of kings and queens ) = 4C2 × 4C2 × 44C1 52C5. To find the probability of obtaining two pairs, we have to consider all possible pairs. Since there are altogether 13 values, that is, aces, deuces, and so on, there are 13 C 2 different combinations of pairs. P( Two pairs ) = 13C2 ⋅ 4C2 × 4C2 × 44C1 52C5 = .04754.

  21. Algebra I Assignment

    This assignment will help students understand the concept of probability and applying statistical concepts to sets of data. Through creating their own data sets, students will practice these skills.

  22. Top Story

    Catch the top stories of the day on ANC's 'Top Story' (18 May 2024)

  23. Federal Register :: Safeguarding and Securing the Open Internet

    Start Preamble Start Printed Page 45404 AGENCY: Federal Communications Commission. ACTION: Final rule. SUMMARY: In this document, the Federal Communications Commission (Commission or FCC) adopts a Declaratory Ruling, Report and Order, Order, and Order on Reconsideration that reestablishes the Commission's authority over broadband internet access service (BIAS).

  24. Combinations, Permutations and Probability Flashcards

    Probability of Simple Events. Teacher 20 terms. Susan_Panasuk. Preview. Stats Final Review- Chapter 13 test. 12 terms. kirbykoster. Preview. Exam_1. 42 terms. jason_perez248. Preview. ISDS 2000 exam 2 shreve. 17 terms. svines1404. Preview. Vocab list #9- module 14/15. 10 terms. quizlette18638500. Preview. STAT 213 - Chapter 4: Probability Part A.