probability Problem

That should be 5C3 / 2^5.



Yes, it is. Do you know what that "5C3" means?

Let's use H for "Heads." Let's use T for "Tails". Every result of five tosses can be written to look like something this: HTTHH. (Here you first get heads, then 2 tails, then 2 more heads.)

000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1 2 33 4 5
Another way to think about it, is that we are choosing different places for the H's. If we choose places 1, 4 & 5, then we get HTTHH. If we choose places 2, 3 & 4, then we get THHHT. The heads are in different positions, which is important.

Every choice of three numbers (like the combination 1, 4 & 5) gives you a different result (like HTTHH). You can choose 3 numbers from 5 numbers in 5C3 different ways. Therefore, you have 5C3 different possible ways to get 3 H's in 5 tosses.



No, that's not correct.

The order of the heads and tails matters.

Here is what doesn't matter: it doesn't matter what order you put the three numbers in. So if you put the heads in places 1, 4 & 5; that is the same as putting them in places 1, 5 & 4. Either way you get HTTHH.

(The red words above are mine.)

Also, your formula is the right one for exactly 3 heads.

It is not correct if you're looking for at least 3 heads.

Whenever I have two heads or more I like to combine them!

http://en.wikipedia.org/wiki/Combination

There are 2 possibilities in each of 5 tosses so there are 2^5 possible outcomes

5HEADS IS 5C5 / 2^5 = 1/32

4HEADS IS 5C4 / 2^5 = 5/32

3HEADS IS 5C3 / 2^5 = 10/32

2HEADS IS 5C2 / 2^5 = 10/32

1 HEAD IS 5C1 / 2^5 = 5/32

0 HEAD IS 5C0/ 2^5 = 1/32

For further reference wiki Permutation