# How many ways are there to win on exactly 3 out of 10 lottery tickets?

This question is for my own understanding of probability.

Suppose that for a particular lottery ticket game, the odds of winning a prize are 1 in 4.66.

I want to know my chances of winning on exactly 3 of my 10 tickets.

I know how to find out the odds of not winning on any of my tickets. I would take the chance of losing (3.66/4.66) to the 10th power and convert this to a percentage, but I am unsure how to determine the probability of 3 out of 10 tickets winning a prize.

The main reason I am confused is because there are multiple ways to win on exactly 3 tickets. This is why I am unsure of what to do.

• Have you heard of a Binomial Distribution? Aug 25 '14 at 20:34
• Why not 10 out of 10? You can give me the money for the 7 winnings you don't want. Aug 25 '14 at 21:30

Hint: Use the Binomial distribution with $p=1/4.66$, $n=10$, and $k=3$.
First figure out how the chances of winning with a particular set of $3$ tickets. As in say you win on your first $3$ tickets $(1/4.66)^3$ and lose on all the others $(3.66/4.66)^7$. Let this probability be $P$. Now it follows that winning on say the $3$rd, $5$th, and $10$th tickets only is also $P$. What we need to figure out now is how many different ways there are to pick $3$ tickets out of the $10$ you have.
There are $10\choose3$ ways to choose which $3$ tickets will win, and each has $P$ chance of occuring.
$$={10\choose3}\cdot{1^3\over4.66^3}\cdot{3.66^7\over4.66^7}$$