Re: Math Question

Paul wrote:

> Actually, what I did was this.  (1-(.95^2))*.05=204 to 1
> It was late and I was tired.  I see now on my notepade
> I jotted down the following: 3c2 * 19...and that's where
> I left it and went to bed.  Almost.


P(exactly two 7's) = 3C2 (1/20)^2 (19/20)^1
P(exactly one 7)  = 3C1 (1/20)^1 (19/20)^2

This formula is from binomial probability,
either the 7 hits or it does not.

Let p = the probability of what you want.
P(k successes in n trials) = nCk p^k (1-p)^(n-k)

P(exactly two 7's) = 3C2 (1/20)^2 (19/20)^1
In the above n = 3, k = 2, p = 1/20.


P(at least two 7's) = P(exactly two 7's) + P(three 7's)

P(at least 7 seven)
= P(exactly one 7) + P(exactly two 7's) + P(three 7's)

You can Google binomial probability.