Question

The probability that a certain NBA player makes a free throw is 77%. In one game, he attempted 12 free throws. What is the probability that he made exactly 3 of them? What is the probability that he made all but one of them? What is the probability that he made more than half of them?

Answer #1

n = 12

p = 0.77

P(X = x) = nCx * p^{x} * (1 - p)^{n - x}

a) P(X = 3) = 12C3 * (0.77)^3 * (0.23)^9 = 0.0002

b) P(X = 1) = 12C1 * (0.77)^1 * (0.23)^11 = 0.00001

c) P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

= 12C7 * (0.77)^7 * (0.23)^5 + 12C8 * (0.77)^8 * (0.23)^4 + 12C9 * (0.77)^9 * (0.23)^3 + 12C10 * (0.77)^10 * (0.23)^2 + 12C11 * (0.77)^11 * (0.23)^1 + 12C12 * (0.77)^12 * (0.23)^0

= 0.9626

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