Question

A basketball player steps to the line to shoot three free-throws. If her free-throw completion average is .89, and we assume her probability of completion is constant and each shot is independent, what is the probability that she completes at least two free-throws? (Round your answer to three decimal places.)

Answer 1

Given , n = 3 , p = 0.89

P(X=x) = nCx px * (1-p)n-x = 3cx 0.89x * (1-0.89)3-x

Therefore, the probability that at least two free-throws is,

P(x>=2) = P(x=2) + P(x=3)

P(x=2) = 3C2 0.892 * (1-0.89)3-2 = 0.261

P(x=3) = 3C3 0.893 * (1-0.89)3-3 = 0.705

P(x>=2) = 0.261 + 0.705 = 0.966