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Step-by-Step Solution
Step 1: Recognize the requirement
We need to form a 4-member committee from 10 men and 5 women, with the condition that each committee has at least one woman. We then want the probability that the committee has more women than men (i.e., either 3 women and 1 man or 4 women and 0 men).
Step 2: Compute the total number of valid committees (at least one woman)
The total number of ways to choose any 4 people out of 15 without restriction is $ ^{15}C_{4} $. However, we only want committees with at least one woman. An effective way is to subtract from $ ^{15}C_{4} $ the committees with no women at all (i.e., all men). But here, we will follow the direct sum of cases:
• Exactly 1 woman and 3 men: $ ^{5}C_{1} \times ^{10}C_{3} $
• Exactly 2 women and 2 men: $ ^{5}C_{2} \times ^{10}C_{2} $
• Exactly 3 women and 1 man: $ ^{5}C_{3} \times ^{10}C_{1} $
• Exactly 4 women and 0 men: $ ^{5}C_{4} $
Hence, the total number of such committees is:
\[
^{5}C_{1} \times ^{10}C_{3}
+ \; ^{5}C_{2} \times ^{10}C_{2}
+ \; ^{5}C_{3} \times ^{10}C_{1}
+ \; ^{5}C_{4}.
\]
Evaluating each term:
\[
^{5}C_{1} = 5,\quad
^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120,
\]
\[
^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10,\quad
^{10}C_{2} = \frac{10 \times 9}{2 \times 1} = 45,
\]
\[
^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10,\quad
^{10}C_{1} = 10,
\]
\[
^{5}C_{4} = 5.
\]
So the sum is:
\[
(5 \times 120) + (10 \times 45) + (10 \times 10) + 5 = 600 + 450 + 100 + 5 = 1155.
\]
Step 3: Compute the number of committees that have more women than men
For a 4-member committee, "more women than men" means either 3 women and 1 man or 4 women and 0 men:
\[
\text{Number of such committees}
= \; ^{5}C_{3} \times ^{10}C_{1} \; + \; ^{5}C_{4}.
\]
We already found:
\[
^{5}C_{3} = 10, \quad ^{10}C_{1} = 10, \quad ^{5}C_{4} = 5.
\]
So,
\[
10 \times 10 + 5 = 100 + 5 = 105.
\]
Step 4: Calculate the probability
The desired probability is the ratio of committees with more women than men to the total committees with at least one woman:
\[
\text{Probability} = \frac{105}{1155} = \frac{1}{11}.
\]
Therefore, the probability that a 4-member committee has more women than men is $ \frac{1}{11} $.
