1. Chapter 8 Counting Principles: Further Probability Topics Section 8.5 Probability Distributions; Expected Value

2. Probability Distributions A table of probabilities sets up a probability distribution. A table of probabilities sets up a probability distribution. Example: A single die is rolled once. Make a probability distribution for all the possible outcomes. Example: A single die is rolled once. Make a probability distribution for all the possible outcomes. OutcomeProbability 11/6 21/6 31/6 41/6 51/6 61/6

3. In other words, for each possible outcome of an experiment, a probability for that outcome is assigned. In other words, for each possible outcome of an experiment, a probability for that outcome is assigned. These possible outcomes represent the idea of a random variable. These possible outcomes represent the idea of a random variable. A random variable is a function that assigns a real number to each outcome of an experiment. A random variable is a function that assigns a real number to each outcome of an experiment. OutcomeProbability 11/6 21/6 31/6 41/6 51/6 61/6

4. A table that lists the possible values of a random variable, together with the corresponding probabilities, is called a probability distribution. A table that lists the possible values of a random variable, together with the corresponding probabilities, is called a probability distribution. The sum of the probabilities of a probability distribution must always equal 1. The sum of the probabilities of a probability distribution must always equal 1. (Rounding may cause this sum to be not exactly 1.)

5. There is just one probability for each value of the random variable (which is basically the definition of a function), thus, we refer to this as a probability distribution function, or probability function. These two terms are used interchangeably. There is just one probability for each value of the random variable (which is basically the definition of a function), thus, we refer to this as a probability distribution function, or probability function. These two terms are used interchangeably.

6. Example: A bag of marbles contains 4 blue, 3 red, 2 white, and 1 green. Three marbles are drawn from the bag at random. What is the probability that the marble(s) in the sample drawn is/are blue? Example: A bag of marbles contains 4 blue, 3 red, 2 white, and 1 green. Three marbles are drawn from the bag at random. What is the probability that the marble(s) in the sample drawn is/are blue? We will create a probability distribution, letting our random variable represent the possible number of blue marbles that can be drawn. We will create a probability distribution, letting our random variable represent the possible number of blue marbles that can be drawn. x = possible number of blue marbles in sample of 3 marbles drawn from bag of 3 marbles drawn from bag x =0, 1, 2, or 3

7. x # of blue marbles P(x)P(x)P(x)P(x) 0 1 2 3 In order to find the corresponding probabilities for each possible outcome, we must use combinations because the order with which we select the marbles for our sample does not matter.

8. The information in a probability distribution is often displayed graphically as a special kind of bar graph called a histogram. The information in a probability distribution is often displayed graphically as a special kind of bar graph called a histogram. The bars of a histogram all have the same width, usually 1, and are adjacent to each other. The bars of a histogram all have the same width, usually 1, and are adjacent to each other. The heights of the bars are determined by the probabilities. The heights of the bars are determined by the probabilities.

9. Create a histogram from the probability distribution for the marbles. (Note: bars should be adjacent.) Create a histogram from the probability distribution for the marbles. (Note: bars should be adjacent.) Chances of Drawing a Blue Marble Probability Number of Blue Marbles Drawn

10. Find the probability that at least two blue marbles are drawn. Find the probability that at least two blue marbles are drawn. Chances of Drawing a Blue Marble Probability Number of Blue Marbles Drawn

11. In working with probability distributions, it’s useful to have a concept of the typical, or average, value that the random variable takes on. This average is called the expected value. In working with probability distributions, it’s useful to have a concept of the typical, or average, value that the random variable takes on. This average is called the expected value. A game with an expected value of 0 is called a fair game.

12. Use the probability distribution below to find the expected number of blue marbles when three marbles are drawn from the bag and interpret the result. Use the probability distribution below to find the expected number of blue marbles when three marbles are drawn from the bag and interpret the result. x0123 P(x).292.525.175.008 E(x) = 0 (.292) + 1 (.525) + 2 (.175) + 3 (.008) =.899 The expected value of.899 means that we should expect to get an average of 1 blue marble every time we draw a marble from the bag.

13. Sometimes, the situation will require the binomial probability formula to be used. Sometimes, the situation will require the binomial probability formula to be used. To find the expected value, we can use the method below. To find the expected value, we can use the method below.

14. A poll conducted by U.S. News and World Report reported that 84% of college students believe they need to cheat to get ahead in the world today. If 500 college students were surveyed, how many would you expect to say that they need to cheat to get ahead in the world today? (Source: Kleiner, Carolyn and Mary Lord, “The Cheating Game,” U.S. News and World Report, Nov. 22, 1999, pp.55-66) A poll conducted by U.S. News and World Report reported that 84% of college students believe they need to cheat to get ahead in the world today. If 500 college students were surveyed, how many would you expect to say that they need to cheat to get ahead in the world today? (Source: Kleiner, Carolyn and Mary Lord, “The Cheating Game,” U.S. News and World Report, Nov. 22, 1999, pp.55-66) n = 500p =.84 n = 500p =.84 E(x) = np = (500) (.84) = 420 E(x) = np = (500) (.84) = 420 420 college students in a group of 500 would be expected to say they needed to cheat based on this study. 420 college students in a group of 500 would be expected to say they needed to cheat based on this study.