To find the probability that both selected people are women, we first need to determine the total number of ways to select two people from the committee and the number of ways to select two women.

Step 1: Calculate the total number of ways to select two people from the committee.

The committee consists of 4 women and 3 men, making a total of:

$4 + 3 = 7 \text{ people}$

The number of ways to select 2 people out of 7 is given by the binomial coefficient:

$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$

Step 2: Calculate the number of ways to select two women from the committee.

The number of ways to select 2 women from the 4 women is:

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

Step 3: Calculate the probability that both selected people are women.

The probability that both selected people are women is the ratio of the number of ways to select two women to the total number of ways to select two people:

$P(\text{both are women}) = \frac{\binom{4}{2}}{\binom{7}{2}} = \frac{6}{21} = \frac{2}{7}$

Final Answer

The probability that both selected people are women is:

$\frac{2}{7}$

Would you like more details on this solution or have any questions?

Follow-up Questions:

1. What is the probability that both selected people are men?
2. What is the probability that one woman and one man are selected?
3. How would the probability change if there were 5 women and 2 men?
4. What is the probability that at least one man is selected?
5. If the committee decided to select three people instead, what is the probability that all three are women?

Tip: When dealing with probabilities, always ensure that the total number of outcomes is correctly calculated to get accurate results.