A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least 1 woman?

### Concept & Formula When calculating "at least one", it is almost always faster to use the complement rule: find the probability of the opposite event ("none") and subtract it from 1. $$ P(\text{at least one woman}) = 1 - P(\text{no women selected}) $$ ### Step-by-Step Solution * **Total people:** 3 men + 2 women = 5 people. * **Total ways to select 3 members:** ${^5C_3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$ ways. * We want committees with at least 1 woman. Let's find the opposite: committees with NO women (meaning all 3 members are men). * **Ways to select 3 men out of 3:** ${^3C_3} = 1$ way. * **Probability of no women:** $\frac{1}{10}$. * **Probability of at least 1 woman:** $1 - P(\text{no women}) = 1 - \frac{1}{10} = \frac{9}{10}$. ### Exam Strategy & Shortcut The complement rule is the ultimate shortcut here. The alternative is adding the probabilities of selecting exactly 1 woman ($^2C_1 \times {^3C_2} = 2 \times 3 = 6$) and exactly 2 women ($^2C_2 \times {^3C_1} = 1 \times 3 = 3$). $6 + 3 = 9$ favorable outcomes. While easy here because numbers are small, this additive method becomes very slow with larger groups. ### Common Pitfall Solving by addition (exactly 1 + exactly 2) but forgetting a case or making an arithmetic error. Relying on the complement rule mitigates these arithmetic risks. ### Final Answer Therefore, the correct answer is **$\frac{9}{10}$**.