Question

A certain tennis player makes a successful first serve 77​% of the time. Assume that each serve is independent of the others. If she serves 4 ​times, what's the probability she gets​ a) all 4 serves​ in? b) exactly 2 serves​ in? c) at least 2 serves​ in? d) no more than 3 serves​ in?

Answer 1

X ~ Bin ( n , p)

Where n = 4 , p = 0.77

Binomial probability distribution is

P(X) = nCx px ( 1 - p)n-x

a)

P(X = 4) = 4C4 * 0.774 * ( 1 - 0.77)0

= 0.3515

b)

P(X = 2) = 4C2 * 0.772 * ( 1 - 0.77)2

= 0.1882

c)

P(X >= 2) = P(X = 2) + P(X = 3) + P(X = 4)

= 4C2 * 0.772 * ( 1 - 0.77)2 + 4C3 * 0.773 * ( 1 - 0.77)1 + 4C4 * 0.774 * ( 1 - 0.77)0

= 0.9597

d)

P(X <= 3) = 1 - P(X = 4)

= 1 - 0.3515

= 0.6485