A committee of $3$ members is to be formed from $5$ men and $4women$. What is the probability that the committee contains at least one woman?

1. Total number of people $=5+4=9$.
2. Total ways to select a $3$ -member committee:

$$n(S)=\left(\genfrac{}{}{0ex}{}{9}{3}\right)=\frac{9\times 8\times 7}{3\times 2\times 1}=84$$

3. Let's find the number of ways to form a committee with NO women (i.e., all men):

$$n(\text{All Men})=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$$

4. Number of ways to select at least one woman:

$$n(E)=84-10=74$$

5. Probability $P(E)=\frac{74}{84}=\frac{37}{42}$.