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December 20, 2014

December 20, 2014

Posted by **Jen** on Monday, February 4, 2008 at 10:55am.

- Statistics -
**economyst**, Monday, February 4, 2008 at 1:37pmThe probability of seeing the first 61 in a batch good and the last 3 defective is (.96)^61 * (.04)^3 = X

Now then, count the number of ways a batch of 64 could have exactly 3 defectives. The formula for n-choose-c or 64-choose-3 is (n!/c!(n-c)!) (where ! means factorial).

64!/3!*61! = 41664.

Finally 41664*X = .221 or 22.1%

QED

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