Question

A certain tennis player makes a successful first serve 61% of the time. Assume that each serve is independent of the others. If she serves 5 times, what's the probability she gets a) all 5 serves in? b) exactly 3 serves in? c) at least 3 serves in? d) no more than 3 serves in?

Answer #1

X: number of serves in.

X ~ bin(n,p)

where, n= 5

p =61% = 0.61

q= (1-p)= (1-0.61) = 0.39

the pmf of the distribution be:-

**a). the probability that she gets all 5 serves in
be:-**

= **0.0845**

**b)the probability that she gets all 5 serves in
be:-**

= **0.3452**

**c).the probability that she gets at least 3 serves in
be:-**

= **0.6997**

d).the probability that she gets no more than 3 serves in be:-

= **0.6455**

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