Asked by Bersy
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.)
Answers
Answered by
Damon
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
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
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