# Math Help - Simple problem...

1. ## Simple problem...

... I'm having trouble with it anyway, even though I think it might just be high school review, so I wasn't entirely sure where to put this topic. Anyway:

There are 15 items. Four are type A, five are type B, and six are type C. Selecting 3 randomly, what is the chance that exactly 2 will be type C?

There is no item replacement and it's unordered.

Since there are more questions using the same sort of thing later I was hoping to get a formula/method I could use with larger numbers instead of tracing through a diagram and multiplying every time.

2. Looking at the categories:

A: 4/15
B: 5/15
C: 6/15

Now, looking at these possible outcome combinations: C,C,N ; C,N,C ; or N,C,C. (N denoting "not C" in this case.)

This would be a simple binominal IF it were done with replacement:

$\binom{n}{k} p^kq^{n-k}$

But since there is no replacement, each succeeding factor is conditional to the previous factor (i.e. you simply remove what you take out at each step):

$Pr(2 C's | 3 Draws) = \binom{3}{2} (6/15)(5/14)(9/13)$

Where 9/13 is the "not C" part, i.e. 4/13 + 5/13