11.10 Binomial Probability Formula 747 In Example 1, we use the binomial probability formula to solve the same problem we recently solved by using a tree diagram. Example 1 Selecting Colored Balls with Replacement A basket contains 3 balls: 1 red, 1 blue, and 1 yellow. Three balls are going to be randomly selected with replacement from the basket. Determine the probability that a) no red balls are selected. b) exactly 1 red ball is selected. c) exactly 2 red balls are selected. d) exactly 3 red balls are selected. Solution a) We will consider selecting a red ball a success and selecting a ball of any other color a failure. Since only 1 of the 3 balls is red, the probability of success on any single trial, p, is . 1 3 The probability of failure on any single trial, q, is 1 . 1 3 2 3 − = We are determining the probability of selecting 0 red balls, or 0 successes. Since x represents the number of successes, we let x 0. = There are 3 independent selections (or trials), so n 3. = In our calculations, we will need to evaluate . 1 3 0 ( ) Note that any nonzero number raised to a power of 0 is 1. Thus, 1. 1 3 0 ( ) = We determine the probability of 0 successes, or P(0), as follows. P x C p q P C ( ) ( ) (0) ( ) 1 3 2 3 (1)(1) 2 3 2 3 8 27 n x x n x 3 0 0 3 0 3 3 = = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ = − − b) We are determining the probability of obtaining exactly 1 red ball or exactly 1 success in 3 independent selections. Thus, x 1 = and n 3. = We determine the probability of exactly 1 success, or P(1), as follows. P x C p q P C ( ) ( ) (1) ( ) 1 3 2 3 3 1 3 2 3 3 1 3 4 9 4 9 n x x n x 3 1 1 3 1 2 = = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = − − c) We are determining the probability of selecting exactly 2 red balls in 3 independent trials. Thus, x 2 = and n 3. = We determine P(2) as follows. P x C p q P C ( ) ( ) (2) ( ) 1 3 2 3 3 1 3 2 3 3 1 9 2 3 2 9 n x x n x 3 2 2 3 2 2 1 = = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = − −
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