Probability question

First answer how many ways to get two failures out of three draws (vombinations)

Call F a bad part draw and G a good part draw since there are so few I'm just going to count

FFG FGF GFF 3 combinations*

now do the with and without replacement probabilities on a single series of draws.

w/o replacement 6/25*5/24*19/23

w replacement 6/25*6/25*19/25

multiply the number of combinations by the probability on a single draw;

Rap

* there's a general routine called combinations that comes from the binomial coefficients http://en.wikipedia.org/wiki/Combination.

In this case it can be calculated by n!/(k!*(n-k)!) where n=3 and k=2