Exhibit 5-13.Chauncey Billups,a current shooting guard for the Los Angeles Clippers,has a career free-throw percentage of 89.4%.Suppose he shoots six free throws in tonight's game. Refer to Exhibit 5-13.What is the probability that Billups makes five or more of his free throws?
A)0.3632
B)0.5105
C)0.8737
D)0.8940

Question

Quizplus · Accepted Answer

This is a binomial probability distribution problem, where the number of trials n = 6 (six free throws), the probability of success (making the free throw) p = 0.894, and we want to find the probability of making 5 or more free throws. Using a binomial probability table or calculator, we can find the probability of making exactly 5 free throws (0.437), making exactly 6 free throws (0.444), and add these probabilities together to get the probability of making 5 or more free throws. 
P(X ≥ 5) = P(X=5) + P(X=6) = 0.437 + 0.444 = 0.881
Rounding to four decimal places, the answer is C) 0.8737.