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Suppose a basketball player makes 80% of her free throws. Let X be the number of...

Suppose a basketball player makes 80% of her free throws. Let X be the number of free throws that she makes in the next n = 10 attempts. Note that X has a binomial distribution with n = 10 and p = 0.8. Find the probability that she makes 6 of her 10 attempts, denoted by P(X = 6).

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Answer #1

P(X=x)=nCx px q(n-x)

n = 10, x = 6 , p = .8 , q = 1-p =.2

P(X= 6) =10C6 (.8)6(.2)4 = 210 * .262*.0016 = 0.088

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