Question

A certain tennis player makes a successful first serve 67% of the time. Assume that each serve is independent of the others. If she serves 7 times, what's the probability she gets a) all 7 serves in? b) exactly 4 serves in? c) at least 5 serves in? d) no more than 4 serves in?

Answer #1

it is given that p = 0.67 and total n =7

using binomial distribution, we will solve this question

(A) All 7 serves in

using binompdf(n,p,k), we have n = 7, p = 0.67 and k = 7

= binompdf(7,0.67,7)

= 0.0606

(B) probability of exactly 4

using binompdf(n,p,k), we have n = 7, p = 0.67 and k = 4

= binompdf(7,0.67,4)

= 0.2535

(C) probability of at least 5 means 1 - P(at most 4)

using binomcdf(n,p,k), where n = 7, p = 0.67 and k = 4

= 1 - P(at most 4)

= 1 - binomcdf(7,0.67,4)

= 1 - 0.4217

= 0.5783

(D) Probability of no more than 4 serves in, this means probability of at most 4

using binomcdf(n,p,k), where n = 7, p = 0.67 and k = 4

= P(at most 4)

= binomcdf(7,0.67,4)

= 0.4217

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