Originally Posted by

**Runty** I need a confirmation on the answer I have here.

A small lake has 105 fish: 40 are trout and 65 are carp. A fisherman caught 8 fish. What is the probability that exactly 2 of them are trout if we know that at least 3 of them are not.

My current answer is as follows:

$\displaystyle P=\frac{C(65,3)\times C(40,5)}{C(105,8)}+\frac{C(65,4)\times C(40,4)}{C(105,8)}+\frac{C(65,5)\times C(40,3)}{C(105,8)}+\frac{C(65,6)\times C(40,2)}{C(105,8)}$

If I did anything wrong, can someone point it out?