1. Interpret the condition

We form a 4-member committee from a group of 10 men and 5 women, with the restriction that the committee must contain at least one woman.

So the sample space consists of all 4-member committees having at least one woman.

2. Count total valid committees

Total number of 4-member committees from 15 people: $\binom{15}{4} = 1365$

Committees with no woman = all men: $\binom{10}{4} = 210$

Hence, total committees with at least one woman: $1365 - 210 = 1155$

3. Count favorable committees

We want committees with more women than men.

Since the committee has 4 members, possible compositions with more women than men are:

• 3 women, 1 man
• 4 women, 0 men

Now count each:

Case 1: 3 women and 1 man

$\binom{5}{3}\binom{10}{1} = 10 \cdot 10 = 100$

Case 2: 4 women and 0 men

$\binom{5}{4}\binom{10}{0} = 5 \cdot 1 = 5$

So total favorable committees: $100 + 5 = 105$

4. Compute the required probability

$P = \frac{105}{1155}$

Simplify: $\frac{105}{1155} = \frac{1}{11}$

5. Match with options

$\frac{1}{11}$ corresponds to Option C.

6. Compare with stored answer

Stored correct answer: C

This matches our derived answer.