A lottery has released n+m tickets for sale, of which only n…
A lottery has released $n+m$ tickets for sale, of which only $n$ are winning tickets. If someone bought $k$ tickets, find the probability that among them there will be exactly $s$ winning tickets.
Explanation of Method
The $k$ bought tickets form a random group of $k$ out of $n+m$, and all groups are equally likely. A favourable group consists of $s$ winning tickets (chosen from the $n$ winning ones) and $k-s$ losing tickets (chosen from the $m$ losing ones). Counting with combinations and dividing gives the hypergeometric probability — the same idea as in Problem 37.
Step 1: All outcomes
\[ \text{number of ways to choose } k \text{ tickets out of } n+m = \binom{n+m}{k}. \]
Step 2: Favourable outcomes
- choose the $s$ winning tickets from the $n$ winning ones: $\binom{n}{s}$ ways;
- choose the other $k-s$ tickets from the $m$ losing ones: $\binom{m}{k-s}$ ways.
By the multiplication rule: $\binom{n}{s}\binom{m}{k-s}$ favourable groups.
Step 3: The probability
\[ \boxed{P = \frac{\dbinom{n}{s}\dbinom{m}{k-s}}{\dbinom{n+m}{k}}, \qquad \max(0,\,k-m)\le s\le\min(k,\,n).} \]
For other values of $s$ there is no favourable group and $P=0$ (you cannot get more winning tickets than exist, or more losing ones than exist).
Step 4: Example
$n+m=20$ tickets, $n=5$ winning, $m=15$ losing; you buy $k=3$. Then $\binom{20}{3}=1140$ and
| $s$ | $\binom5s\binom{15}{3-s}$ | Probability |
|---|---|---|
| $0$ | $1\cdot455=455$ | $455/1140\approx0.399$ |
| $1$ | $5\cdot105=525$ | $525/1140\approx0.461$ |
| $2$ | $10\cdot15=150$ | $150/1140\approx0.132$ |
| $3$ | $10\cdot1=10$ | $10/1140\approx0.009$ |
| total | $1140$ | $1$ ✓ |
So the chance of exactly one winning ticket is about $0.46$, and the chance of at least one is $1-0.399=0.601$.
Step 5: Average number of winning tickets
Each bought ticket wins with probability $\frac{n}{n+m}$, so on average $k\cdot\frac{n}{n+m}$ of the bought tickets win (in the example $3\cdot\frac{5}{20}=0.75$).
Result
| Answer | |
|---|---|
| Probability of exactly $s$ winning tickets | $P=\dfrac{\binom{n}{s}\binom{m}{k-s}}{\binom{n+m}{k}}$ |
| Possible values | $\max(0,k-m)\le s\le\min(k,n)$ |