Originally Posted by

**Jasonvdm** Hey guys, here's my problem:

"Ten percent of cadets at the police academy chew gum regularly. What is the probability (to the nearest percent) that, if you pick 4 cadets at random, two of them chew gum regularly?"

Well, that was easy. You use the binomial distribution. Part 2 of the question gets tricky though.

"If you 800 of them, what is the probability that 83 chew gum regularly? (answer is more than 37%)"

Here is my equation:

800 C 83 * 0.10^83(1 - 0.10)^717

Some pretty big, and small numbers that make my calculator hate me.

The permutation isn't hard to understand, just big. I understand that you will only have 800! to 718!, but there's got to be a way of computing that other than long hand writing it down.

Also, what about the tiny numbers?

thanks!