In a lottery $40$ tickets are sold and there are $3$ prizes. What is the probability that a girl buying $5$ tickets wins at least $1$ prize?
Answer is $667/1976$, how do you get there?
I tried finding the probability of getting exactly $1$, $2$ and $3$ prizes but I still don't know how to do it. Any help?
[Math] Ticket Prize Probability
probability
Best Answer
First we do it the way you proposed. For this problem, it is a little inefficient, but the idea will be needed at other times.
There are $\binom{40}{5}$ ways to choose $5$ tickets from $40$.
We find the number of $5$-ticket hands that win exactly $1$ prize. There are $\binom{3}{1}$ ways to choose the "good" ticket, and for each of these ways there are $\binom{37}{4}$ ways to choose the "bad" tickets, for a total of $\binom{3}{1}\binom{37}{4}$ one prize hands.
Thus the probability of winning exactly one prize is $\dfrac{\binom{3}{1}\binom{37}{4}}{\binom{40}{5}}$.
Similarly, the number of two prize hands is $\binom{3}{2}\binom{37}{3}$.
Thus the probability of winning exactly two prizes is $\dfrac{\binom{3}{2}\binom{37}{3}}{\binom{40}{5}}$.
Finally, the probability of winning exactly three prizes is $\dfrac{\binom{3}{3}\binom{37}{2}}{\binom{40}{5}}$.
Add up.
Another way: We find the probability $p$ of winning no prize. Then the probability of winning at least one is $1-p$.
There are $\binom{37}{5}$ hands in which all the tickets are bad. It follows that $$p=\dfrac{\binom{37}{5}}{\binom{40}{5}}.$$
Still another way: Again, we first find the probability of no prize. Suppose she checks her tickets one by one to see whether she has won something.
The probability the first ticket is bad is $\frac{37}{40}$. Given that the first ticket was bad, the probability the second ticket is bad is $\frac{36}{39}$, for there are only $36$ bads left. So the probability the first two tickets are bad is $\frac{37}{40}\cdot \frac{36}{39}$. Given the first two were bad, the probability the third is bad is $\frac{35}{38}$. And so on. Thus $$p=\frac{37}{40}\cdot \frac{36}{39}\cdot \frac{35}{38}\cdot \frac{34}{37}\cdot \frac{33}{36}.$$ To evaluate, it can be useful to do some cancellation.