Drawing a hand of cards or choosing a committee are draws without replacement: once someone is chosen, they cannot be chosen again, so the draws are not independent and the binomial model does not apply. Counting equally likely subsets, as in probability with counting, gives the exact pmf.
Key idea
Draw $n$ items at random, without replacement, from $N$ items of which $K$ are "successes". The number of successes $X$ has the hypergeometric distribution: $$P(X = k) = \frac{\binom{K}{k}\binom{N - K}{n - k}}{\binom{N}{n}}, \qquad E[X] = n\frac{K}{N}.$$
Why it works
All $\binom{N}{n}$ sets of $n$ items are equally likely. A set with exactly $k$ successes is made by choosing $k$ of the $K$ successes and $n - k$ of the $N - K$ others: $\binom{K}{k}\binom{N - K}{n - k}$ ways. For the mean, use indicators: by symmetry, draw $j$ is equally likely to be any of the $N$ items, so it is a success with probability $\frac{K}{N}$, and the $n$ draws add to $n\frac{K}{N}$.
Worked examples
Worked example: A committee
A committee of 4 is chosen at random from 7 women and 5 men. Find the probability that it has exactly 2 women, and the expected number of women.
- $$P(X = 2) = \dfrac{\binom{7}{2}\binom{5}{2}}{\binom{12}{4}} = \dfrac{21 \cdot 10}{495} = \frac{14}{33} \approx 0.424$$
- $E[X] = 4 \cdot \frac{7}{12} = \frac{7}{3}$.
Answer: $\frac{14}{33}$ and $\frac{7}{3}$
Worked example: At least one defective
A box of 12 bulbs has 3 defective ones. You test 4 at random. Find the probability that at least one is defective.
- Complement: none defective, all 4 from the 9 good bulbs: $\dfrac{\binom{9}{4}}{\binom{12}{4}} = \dfrac{126}{495} = \frac{14}{55}$.
- $P(X \ge 1) = 1 - \frac{14}{55} = \frac{41}{55} \approx 0.745$.
Answer: $\frac{41}{55}$
Worked example: When the binomial is close
Compare $P(2 \text{ women})$ in the committee example with the binomial value for 4 independent draws with $p = \frac{7}{12}$, and with a group of 700 women and 500 men.
- Binomial: $\binom{4}{2}\left(\frac{7}{12}\right)^{2}\left(\frac{5}{12}\right)^{2} \approx 0.354$, quite different from $0.424$.
- With 700 women and 500 men the hypergeometric value is about $0.355$: removing a few people barely changes the mix, so the draws are nearly independent.
Answer: The binomial is a good approximation only when $N$ is large compared with $n$.
Common mistake
Using the binomial for draws without replacement from a small group. After one woman is chosen, the chance that the next is a woman drops from $\frac{7}{12}$ to $\frac{6}{11}$, so the trials are neither independent nor identical. Check: if $n$ is more than about a tenth of $N$, use the hypergeometric.