Posted by Bersy on Sunday, January 27, 2013 at 3:30pm.
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
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