8.5 Probability Distributions; Expected Value

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1 Math 07 - Finite Math.5 Probability Distributions; Expected Value In this section, we shall see that the expected value of a probability distribution is a type of average. A probability distribution depends on the idea of a random variable, so we begin with that. Random Variables We will begin with an example. Example. Suppose a coin is tossed 3 times and the sequence of heads and tails is noted. The sample space for the experiment is S = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }. Now, we label each outcome by the number of heads in the outcome. That is, we assign to each outcome in S a number from the set {0,,, 3}. The table below lists the outcomes of S and their corresponding labels. We call this labeling x. Outcome HHH HHT HT H HT T T HH T HT T T H T T T Label (x) 3 0 Then we can think of this labeling as a function from outcomes to numbers. The function x is called a random variable. Random Variable A random variable is a function that assigns a real number to each outcome of an experiment. Probability Distributions In the example of flipping a coin 3 times, we can calculate the probability that 0,,, or 3 heads show up. Example. For each of the following values of x, find the probability that x heads show up. (a) x = 0 Solution. Flipping a coin 3 times is a binomial experiment as we saw in Section.. So using the formula for binomial probability, (b) x = P (0 heads) = ( ) 3 0 ( ) 0 ( ) 3 = =. Fall 00 Page Penn State University

2 Math 07 - Finite Math Solution. P ( head) = ( ) 3 ( ) ( ) = 3 = 3. (c) x = Solution. P ( heads) = ( ) 3 ( ) ( ) = 3 = 3. (d) x = 3 Solution. P (3 heads) = ( ) 3 3 ( ) 3 ( ) 0 = =. Now we can arrange these possible values of x along with their probabilities in a table as shown below. x 0 3 P (x) 3 3 A table as in the above example that lists all the possible values of the random variable, together with the corresponding probabilities, is called a probability distribution. Following are some properties of a probability distribution. The sum of the probabilities in a probability distribution must always equal. (The sum in some distributions may slightly vary from because of rounding.) There is just one probability for each value of the random variable. Thus a probability distribution defines a function, called a probability distribution function or simply a probability function. Remark. Note that the probability function and the random variable are different functions: the random variable assigns a number to each outcome of the sample space for an experiment, while the probability function assigns a probability to each value of the random variable. The data in a probability distribution is often displayed as a bar graph called a histogram. Following are the properties of such a histogram. There is a bar for each value of the random variable. The bars all have the same width, usually unit. Fall 00 Page Penn State University

3 Math 07 - Finite Math The height of a bar is determined by the probability of the value of the random variable represented by the bar. Example 3. Draw a histogram for the probability distribution in Example. Solution. The following table gives the probability distribution in Example. x 0 3 P (x) 3 3 The following figure shows the histogram for the above probability distribution. Example. (a) Give the probability distribution for the number of heads showing when two coins are tossed. Solution. Let x be the random variable, which represents the number of heads. Then x can take on the values 0,, or. Now we need to find the probability of each outcome. To find the probability of 0,, or heads, we can either use binomial probability, or note that there are equally likely outcomes in the sample space: {HH, HT, T H, T T }. The results are shown in the following table, which gives the probability distribution for the number of heads showing. x 0 P (x) Fall 00 Page 3 Penn State University

4 Math 07 - Finite Math (b) Draw a histogram for the distribution and shade the region corresponding to the event at least one head showing. What is the probability of at least one head showing? Solution. The figure below gives the histogram for the above probability distribution. The region shaded in green represents P (x ) = P (x = ) + P (x = ) = + = 3. Example 5. Suppose that the shipping manager at a company receives a package of one dozen computer monitors, of which, unknown to him, three are broken. He checks four of the monitors at random to see how many are broken in his sample of. Find the probability distribution for the number of broken monitors in the sample of. Solution. Let x be the random variable, which represents the number of broken monitors in the sample of. Thus, the possible values of the random variable are 0,,, and 3 (because there are only 3 broken monitors in the package). Now we need to find the probabilities of having 0,,, or 3 broken monitors in the sample. x = 0 The number of ways of selecting 0 broken monitors from the 3 in the package, and hence good monitors from among the 9 good monitors in the package ( ) ( ) 3 9 = = 6 = 6. 0 The number of ways of selecting any monitors from the package ( ) = = 95. Thus, the probability of selecting 0 broken monitors = P (x = 0) = 6 95 = Fall 00 Page Penn State University

5 Math 07 - Finite Math x = Similarly, we can find P (x = ). P (x = ) = ( ) ( ) ( ) = 3 95 =. x = x = 3 ( ) ( ) 3 9 P (x = ) = ( ) = P (x = 3) = =. ( ) ( ) ( ) = 9 95 =. Thus, the probability distribution for the number of broken monitors is x 0 3 P (x) The histogram for the above probability distribution is shown below. Fall 00 Page 5 Penn State University

6 Math 07 - Finite Math Expected Value In working with probability distributions, it is useful to have a concept of the typical or average value that the random variable takes on. In Example, for instance, it seems reasonable that, on an average, one head shows when two coins are tossed. This does not tell what will happen the next time we toss two coins; we may get two heads, one head, or none. If we tossed two coins many times, however, we would expect that, in the long run, we would average about one head for each toss of two coins. A way to solve this problem is to imagine flipping the two coins times. Based on the probability distribution in Example, we would expect that out of the times we would get 0 heads, out of the times we would get head, and out of the times we would get heads. The total number of heads we would get, then, is =. The expected number of heads per toss is found by dividing the total number of heads by the total number of tosses, or = =. Notice that the expected number of heads turns out to be the sum of the three values of the random variable x multiplied by their corresponding probabilities. We can use this idea to define the expected value of a random variable as follows. Expected Value Suppose the random variable variable x can take on the n values x, x, x 3,..., x n. Also, suppose the probabilities that these values occur are, respectively, p, p, p 3,..., p n. (That is, P (x = x ) = p, P (x = x ) = p, P (x = x 3 ) = p 3,..., P (x = x n ) = p n.) Then the expected value of the random variable is E(x) = x p + x p + x 3 p x n p n. Example 6. In the example with the computer monitors (Example 5), find the expected number of broken monitors that the shipping manager finds in his sample of monitors. Solution. The following table gives the probability distribution of the random variable x representing the number of broken monitors in a sample. x 0 3 P (x) Fall 00 Page 6 Penn State University

7 Math 07 - Finite Math Then using the definition of expected value, we have E(x) = =. Thus, on the average, the shipping manager will find broken monitor in the sample of. On reflection, this seems natural; 3 out of the monitors, or of the total, are broken. We should expect, that of the sample of monitors are broken. Example 7. Suppose a local symphony decides to raise money by raffling a microwave oven worth $00, a dinner for two worth $0, and books worth $0 each. A total of 000 tickets are sold at $ each. Find the expected payback for a person who buys one ticket in the raffle. Solution. Here the random variable represents the possible amounts of payback, where payback = amount won cost of ticket. The payback of the person winning the oven is $00 (amount won) $ (cost of ticket) = $399, the payback for each losing ticket is $0 $ = $, etc. The paybacks of the various prizes, as well as their respective probabilities, are shown in the table below. The probability of winning $399 is, because there is only one prize 000 worth $00, and there are 000 tickets. Similarly, the probability of winning a dinner for two worth $0 (payback = $79) is. The probability of winning a book worth $0 (payback 000 = $9) is, as there are such prizes. Finally, since there are winning tickets, there 000 are 996 losing tickets, so the probability of winning $ is x $399 $79 $9 $ P (x) /000 /000 / /000 Hence, the expected payback for a person buying one ticket is E(x) = ( ) = = = 0.7. Thus, on the average, a person buying one ticket in the raffle will lose $0.7, or 7. Note that, it is not possible to lose 7 in this raffle: either you lose $, or you win a prize worth $00, $0, or $0, minus the $ you pay to play. But if you bought tickets in many such raffles over a long period of time, you would lose 7 per ticket on the average. It is important to note that the expected value of a random variable may be number that can never occur in any one trial of the experiment. Fall 00 Page 7 Penn State University

8 Math 07 - Finite Math Example. Each day Donna and Mary toss a coin to see who buys coffee ($.0 a cup). One tosses and the other one calls the outcome. If the person who calls the outcome is correct, the other buys the coffee; otherwise the caller pays. Find Donna s expected winnings. Solution. We assume that an honest coin is used, that Mary tosses the coin, and that Donna calls the outcome. The possible results and corresponding probabilities are shown below. Possible Results Result of Toss Heads Heads Tails Tails Call Heads Tails Heads Tails Caller Wins? Yes No No Yes Probability / / / / Donna wins a $.0 cup of coffee whenever the results and calls match, and she loses a $.0 cup when there is no match. From the above table, we note that the probability that Donna wins is, the same as the probability of her losing the bet. Let x be the random variable, which represents Donna s winnings. So, the possible values of x are $.0 and $.0. Hence, we have the following probability distribution. Thus, Donna s expected winnings are x $.0 $.0 P (x) / / E(x) =.0 + (.0) = $0. On the average, over the long run, Donna neither wins nor loses. Remark. A game with an expected value of 0 (such as in the above example) is called a fair game. Casinos do not offer fair games. If they did, they would win (on the average) $0, and have a hard time paying the help! Example 9. At age 50, you receive a letter from the insurance company. According to the letter, you must tell the company which of the following options you will choose: take $0,000 at age 60 (if you are alive, $0 otherwise), or $30,000 at age 70 (again, if you are alive, $0 otherwise). Based only on the idea of expected value, which should you choose? Solution. Life insurance companies have constructed elaborate tables showing the probability of a person living a given number of years into the future. From a recent such table, the probability of living from age 50 to 60 is 0., while the probability of living from age 50 to 70 is 0.6. The random variable represents your payoff. The probability distributions for the two options are given in the following two tables. Option : x P (x) = 0. Fall 00 Page Penn State University

9 Math 07 - Finite Math So, the expected value of the random variable (payoff) is Option : E(x) = = x P (x) = 0.36 So, the expected value of the random variable (payoff) is E(x) = = 900. Thus, based strictly on the expected values, the second option is preferable. Example 0. According to the National Center for Education Statistics, 7.5% of those earning bachelor s degrees in education in the United States in were female. Suppose 5 holders of bachelor s degrees in education from 003 to 00 are picked at random. (a) Find the probability distribution for the number that are female. Solution. We first note that each of the 5 people in the sample is either female (with probability 0.75) or male (with probability 0.5). As in the previous section, we may assume that the probability for each member of the sample is independent of that of any other. Such a situation is described by binomial probability with n = 5 and p = 0.75, for which we use the binomial probability formula ( ) n p x ( p) n x, x where x is the number of females in the sample (the random variable). For example, ( ) 5 P (x = 0) = (0.75) 0 (0.5) Similarly, we can calculate the probability that x is any value from 0 to 5, resulting in the following probability distribution. x P (x) (b) Find the expected number of females in the sample of 5 people. Solution. Using the formula for expected value, we have E(x) = = 3.95 On the average, 3.95 of the people in the sample of 5 will be female. Fall 00 Page 9 Penn State University

10 Math 07 - Finite Math Remark. There is another way to get the answer in part (b) of the above example. Because 7.5% of those earning bachelor s degrees in education in the United States in are female, it is reasonable to expect 7.5% of our sample to be female. Thus 7.5% of 5 is = Notice that what we have done is to multiply n by p. It can be shown that this method always gives the expected value for binomial probability. Expected Value for Binomial Probability For binomial probability, E(x) = np, where n = number of trials, and p = the probability of one success. In other words, the expected number of successes is the number of trials times the probability of success in each trial. Example. Suppose a family has 3 children. (a) Find the probability distribution for the number of girls. Solution. Assuming girls and boys are equally likely, the probability distribution is binomial with n = 3 and p =. Let the random variable be x, which represents the number of successes (girls). Then using the formula for binomial probability, we find, for example, P (x = 0) = ( ) 3 0 ( ) 0 ( ) 3 =. The other values are found similarly, and the results are shown in the following table. x 0 3 P (x) / 3/ 3/ / We can verify this by noticing that in the sample space S of all 3-child families, there are eight equally likely outcomes: S = {ggg, ggb, gbg, gbb, bgg, bgb, bbg, bbb}. One of the outcomes has 0 girls, three have girl, three have girls, and one has 3 girls. (b) Find the expected number of girls in a 3-child family using the distribution from part (a). Solution. Using the formula for expected value, we have E(x) = Expected number of girls = = =.5. On the average, a 3-child family will have.5 girls. The result agrees with our intuition that, on the average, half of the children born will be girls. Fall 00 Page 0 Penn State University

11 Math 07 - Finite Math (c) Find the expected number of girls in a 3-child family using the formula for expected value for binomial probability. Solution. Using the formula E(x) = np with n = 3 and p =, we have E(x) = Expected number of girls = 3 =.5. This agrees with our answer from part (b), as it must. Fall 00 Page Penn State University