1. Copyright © Cengage Learning. All rights reserved. 8 PROBABILITY DISTRIBUTIONS AND STATISTICS

2. Copyright © Cengage Learning. All rights reserved. 8.4 The Binomial Distribution

3. 3 Probabilities in Bernoulli Trials

4. 4 Let’s reexamine the computations we performed in the last example. There, it was found that the probability of obtaining exactly one success in a binomial experiment with four independent trials with probability of success in a single trial p is given by P(E) = 4pq 3 Observe that the coefficient 4 of pq 3 appearing in Equation (11) is precisely the number of outcomes of the experiment with exactly one success and three failures, the outcomes being SFFF FSFF FFSF FFFS (where q = 1 – p) (11)

5. 5 Probabilities in Bernoulli Trials Another way of obtaining this coefficient is to think of the outcomes as arrangements of the letters S and F. Then the number of ways of selecting one position for S from four possibilities is given by C (4, 1) = = 4

6. 6 Probabilities in Bernoulli Trials Next, observe that because the trials are independent, each of the four outcomes of the experiment has the same probability, given by pq 3 where the exponents 1 and 3 of p and q, respectively, correspond to exactly one success and three failures in the trials that make up each outcome. As a result of the foregoing discussion, we may write Equation (11) as P(E) = C(4, 1)pq 3 (12)

7. 7 Probabilities in Bernoulli Trials We are also in a position to generalize this result. Suppose that in a binomial experiment the probability of success in any trial is p. What is the probability of obtaining exactly x successes in n independent trials? We start by counting the number of outcomes of the experiment, each of which has exactly x successes.

8. 8 Probabilities in Bernoulli Trials Now, one such outcome involves x successive successes followed by (n – x) failures—that is, SS... S FF... F The other outcomes, each of which has exactly x successes, are obtained by rearranging the S’s (x of them) and F’s (n – x of them). There are C(n, x) ways of arranging these letters. (13)

9. 9 Probabilities in Bernoulli Trials Next, arguing as in Example 1, we see that each such outcome has probability given by p x q n – x For example, for the outcome (13), we find P(SS... S FF... F) = P(S)P(S)... P(S) P(F)P(F)...P(F) = pp... p qq... q = p x q n – x Let’s now state this important result formally.

10. 10 Probabilities in Bernoulli Trials If we let X be the random variable that gives the number of successes in a binomial experiment, then the probability of exactly x successes in n independent trials may be written P(X = x) = C(n, x)p x q n – x (x = 0, 1, 2,..., n)(14)

11. 11 Probabilities in Bernoulli Trials The random variable X is called a binomial random variable, and the probability distribution of X is called a binomial distribution.

12. 12 Example 2 A fair die is rolled five times. If a 1 or a 6 lands uppermost in a trial, then the throw is considered a success. Otherwise, the throw is considered a failure. a. Find the probabilities of obtaining exactly 0, 1, 2, 3, 4, and 5 successes in this experiment. b. Using the results obtained in the solution to part (a), construct the binomial distribution for this experiment, and draw the histogram associated with it.

13. 13 Example 2(a) – Solution This is a binomial experiment with X, the binomial random variable, taking on each of the values 0, 1, 2, 3, 4, and 5 corresponding to exactly 0, 1, 2, 3, 4, and 5 successes, respectively, in five trials. Since the die is fair, the probability of a 1 or a 6 landing uppermost in any trial is given by p = =, from which it also follows that q = 1 – p =.

14. 14 Example 2(a) – Solution Finally, n = 5, since there are five trials (throws of the die) in this experiment. Using Equation (14), we find that the required probabilities are cont’d

15. 15 Example 2(a) – Solution cont’d

16. 16 Example 2(b) – Solution cont’d Using these results, we find the required binomial distribution associated with this experiment given in Table 12. Table 12

17. 17 Example 2(b) – Solution Next, we use this table to construct the histogram associated with the probability distribution (Figure 12). cont’d Figure 12 The probability of the number of successes in five throws

18. 18 Probabilities in Bernoulli Trials The following formulas (which we state without proof ) will be useful in solving problems that involve binomial experiments.

19. 19 Probabilities in Bernoulli Trials

20. 20 Probabilities in Bernoulli Trials

21. 21 Probabilities in Bernoulli Trials

22. 22 Example 4 – Solution For the experiment in Example 2 compute the mean, the variance, and the standard deviation of X by (a) using Equations (15a), (15b), and (15c) and (b) using the definition of each term. Solution: a. We use Equations (15a), (15b), and (15c), with p =, q = and n = 5, obtaining

23. 23 Example 4 – Solution We leave it to you to interpret the results. cont’d

24. 24 Example 4 – Solution b. Using the definition of expected value and the values of the probability distribution shown in Table 12, we find that  = E(X)  (0)(.132) + (1)(.329) + (2)(.329) + (3)(.165) + (4)(.041) + (5)(.004)  1.67 which agrees with the result obtained in part (a). cont’d

25. 25 Example 4 – Solution Next, using the definition of variance and  = 1.67, we find that Var (X) = (.132) (–1.67) 2 + (.329) (–0.67) 2 + (.329) (0.33) 2 + (.165) (1.33) 2 + (.041) (2.33) 2 + (.004) (3.33) 2  1.11  X =   1.05 which again agrees with the preceding results. cont’d

26. 26 Practice p. 472 Self-Check Exercises #2