A man throws a basket ball in a hoop to win a Kewpie doll. His probability of winning on each throw is 0.1. The man has three children, and he must win a Kewpie doll for each one.

What is the probability that 10 throws will be required to win the three dolls?

I thought the answer was (.9)^7(.1)^3 although this is not the answer in the book. It says the answer is 0.01722. What am I doing wrong?

See, this is the kind of problem I would freak out on while I was learning this stuff.

The thing is like this...

Let's "code" how the man shoots like this:
X - he won
O - he missed

So, if you write something like OOOOOOOXXX that means he missed the first 7 shots, but he then shot the next three right. This is one way for him to win the 3 dolls.

But what if he shot like this: XXOOOOOOOX ? He would have again won the 3 dolls. XOOOOXOOOX is also one way of winning it. So the question is: In how many ways can he win the 3 dolls in 10 throws?

In order to need EXACTLY 10 shots to win 3 dolls, it means he must win the third doll on his 10th shot. So X must be on the 10th place. That means that we can put 2 Xs and 7 Os in whatever order we want to (before the last X). So the equivalent question to that above is: In how many ways can we put 2 Xs and 7 Os in line?

It's quite easy. You have to choose 2 places for Xs, and the rest are Os. So, you have to choose 2 out of 9 places (it's a binom coefficient "9 over 2") and you can do that in 36 ways. So your answer would be (.9)^7*(.1)^3 * 36 (cuz he can win it in 36 different ways) which makes 0.01722.

Thanks a lot. That makes sense.