A lottery has n tickets, of which m are winning tickets.…

A lottery has $n$ tickets, of which $m$ are winning tickets. What is the probability that someone who bought $k$ tickets will win?

Step 1: All outcomes

\[ \binom{n}{k} \text{ equally likely groups of } k \text{ tickets}. \]

Step 2: The opposite event — no winning ticket

All $k$ tickets come from the $n-m$ losing tickets:

\[ P(\text{no win}) = \frac{\binom{n-m}{k}}{\binom{n}{k}}. \]

(If $k > n-m$, there are not enough losing tickets, $\binom{n-m}{k}=0$, and a win is certain.)

Step 3: The answer

\[ \boxed{P(\text{win}) = 1 - \frac{\binom{n-m}{k}}{\binom{n}{k}}.} \]

Step 4: A product form

Imagine taking the tickets one by one. The first is losing with probability $\frac{n-m}{n}$, then the second with $\frac{n-m-1}{n-1}$, and so on:

\[ P(\text{no win}) = \frac{n-m}{n}\cdot\frac{n-m-1}{n-1}\cdots\frac{n-m-k+1}{n-k+1}, \]

which equals $\binom{n-m}{k}/\binom nk$ after cancelling factorials.

Step 5: Example

$n=100$ tickets, $m=10$ winning, $k=5$ bought:

\[ P(\text{no win}) = \frac{90}{100}\cdot\frac{89}{99}\cdot\frac{88}{98}\cdot\frac{87}{97}\cdot\frac{86}{96} \approx 0.5838, \qquad P(\text{win}) \approx 0.4162. \]

Note: this is a bit more than $1-0.9^5\approx0.410$, because after drawing a losing ticket the remaining tickets contain a slightly larger share of winners. For a very large lottery the two numbers are practically equal.