11.10 Binomial Probability Formula 745 Our three selections may yield 0, 1, 2, or 3 red balls. We can determine the probability of selecting exactly 0, 1, 2, or 3 red balls by using the sample space. To determine the probability of selecting 0 red balls, we count those outcomes that do not contain a red ball. There are 8 of them bbb bby byb byy ybb yby yyb yyy ( , , , , , , , ). Thus, the probability of obtaining exactly 0 red balls is 8/27. We determine the probability of selecting exactly 1 red ball by counting the sample points that contain exactly 1 red ball. There are 12 of them. Thus, the probability is , 12 27 or . 4 9 We can determine the probability of selecting exactly 2 red balls and exactly 3 red balls in a similar manner. The probabilities of selecting exactly 0, 1, 2, and 3 red balls are illustrated in Table 11.5. Table 11.5 A Probability Distribution for Three Balls Selected with Replacement Number of Red Balls Selected, x( ) Probability of Selecting the Number of Red Balls, P x( ) 0 8 27 1 12 27 2 6 27 3 1 27 Sum 27 27 1 = = Note that the sum of the probabilities in Table 11.5 is 1. This table is an example of a probability distribution, which shows the probabilities associated with each specific outcome of an experiment. In a probability distribution, every possible outcome must be listed and the sum of the probabilities must be 1. Let us specifically consider the probability of selecting 1 red ball in 3 selections. We see from Table 11.5 that this probability is 12 27, or . 4 9 Can we determine this probability without developing a tree diagram? The answer is yes. Suppose that we consider selecting a red ball success, S, and selecting a non–red ball failure, F. Furthermore, suppose that we let p represent the probability of success and q the probability of failure on any trial. Then p 1 3 = and q . 2 3 = We can obtain 1 success in 3 selections in the following ways: SFF FSF FFS We can compute the probabilities of each of these outcomes using the multiplication formula because each of the selections is independent since the selections are done with replacement. P P P P p q q pq P P P P q p q pq P P P P q q p pq (SFF) (S) (F) (F) 1 3 2 3 4 27 (FSF) (F) (S) (F) 1 3 2 3 4 27 (FFS) (F) (F) (S) 1 3 2 3 4 27 Sum 12 27 4 9 2 2 2 2 2 2 = ⋅ ⋅ = ⋅ ⋅ = = ⎛ ⎝ ⎞ ⎠ = = ⋅ ⋅ = ⋅ ⋅ = = ⎛ ⎝ ⎞ ⎠ = = ⋅ ⋅ = ⋅ ⋅ = = ⎛ ⎝ ⎞ ⎠ = = =
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