From a group of 3 men and 2 women, a committee of 3 is to be selected uniformly at random. What is the probability that the committee has at least one woman?

Introduction / Context:
“At least one” events are often faster via the complement rule: P(at least one woman) = 1 − P(no women). Counting committees uses combinations since only membership matters, not order.

Given Data / Assumptions:

• Total people = 3 men + 2 women = 5.
• Committee size = 3.
• All committees are equally likely.

Concept / Approach:

• Total committees = C(5,3).
• Committees with no women = choose all from the 3 men: C(3,3).
• Probability = 1 − [C(3,3)/C(5,3)].

Step-by-Step Solution:

Verification / Alternative check:
Direct count of “at least one woman” committees = total 10 minus the single all-men committee = 9 favorable; probability 9/10.

Why Other Options Are Wrong:

• 1/10 is the complement (no woman), not the wanted event.
• 9/20 and 1/20 come from incorrect denominators or double counting.

Common Pitfalls:

• Overcounting when splitting into cases (exactly one, two, or three women) instead of using the quick complement.

Final Answer:
9/10