If the probability that a certain tennis player will serve an ace is 1/4, what is the probability that he will serve exactly two aces out of four serves? (Assume that the four serves are independent. Round your answer to four decimal places.)

Answer 1

I guess I will have to use binomial coefficients.

the probability of k successes in n trials is:

P(k) = C(n,k) p^k (1-p)^(n-k)
C(n,k) is binomial coef
get from Pascal triangle or table or calculate from
C(n,k) - n! / [ k! (n-k)! ]

here
p = prob of ace = .25
(1-p) = .75
n = 4
k = 2

C(4,2) = 4! /[2!(2!)] = 4*3*2/[2(2)] = 6
so
P(2) = 6 (.25)^2 (.75)^2
= .21