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October 30, 2014

October 30, 2014

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

- Math -
**Damon**, Sunday, January 27, 2013 at 4:19pmI 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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