Picking a committee with exactly two men

Question

A student club has 7 women and 5 men. To run an event, they draw 4 names at random from a hat containing all 12 members, every group of four is equally likely.

What is the probability the chosen committee has exactly two men (and therefore two women)?

Accepted Answer

Because order does not matter, count combinations. The total number of four-person committees is \binom{12}{4} = 495. For a committee with exactly two men and two women, choose the men and women separately, then multiply: \binom{5}{2}\binom{7}{2} = 10 	imes 21 = 210. So the probability is \frac{210}{495} = \frac{14}{33} \approx 0.424. Takeaway. When you need a specific composition from two pools, choose from each pool independently and multiply the counts, this is the hypergeometric pattern of sampling without replacement. Divide by the total \binom{n}{k} to turn counts into a probability.

Picking a committee with exactly two men

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