Question

From a group of 10 men and 5 women, a four-member committee which includes at least one woman is to be formed. Then the probability for the committee thus formed to have more women than men is:

• A. \( \frac{3}{11} \)
• B. \( \frac{2}{23} \)
• C. \( \frac{1}{11} \)
• D. \( \frac{21}{220} \)

Hint:

Always read probability conditions carefully. If the phrasing implies choosing from the restricted set containing at least one woman, the sample space size must be reduced from the total global combination count.

Answer 1

The Correct Option is C

Concept: The problem asks for a conditional probability: finding the probability that there are more women than men in a committee of 4, given that the committee contains at least one woman. \[ P = \frac{\text{Number of ways to choose more women than men}}{\text{Total number of ways to choose a committee with at least one woman}} \]

Step 1: Finding the denominator (Total valid committee options).
The total number of ways to choose any 4 members out of 15 is \( \binom{15}{4} \). The number of ways to choose a committee with no women (only men) is \( \binom{10}{4} \). \[ \text{Denominator} = \binom{15}{4} - \binom{10}{4} = \frac{15 \times 14 \times 13 \times 12}{24} - \frac{10 \times 9 \times 8 \times 7}{24} \] \[ = 1365 - 210 = 1155 \]

Step 2: Finding the numerator (More women than men).
For a 4-member committee to have more women than men, the composition must be:

• Case 1: 3 women and 1 man. \[ \text{Ways} = \binom{5}{3} \times \binom{10}{1} = 10 \times 10 = 100 \]

• Case 2: 4 women and 0 men. \[ \text{Ways} = \binom{5}{4} \times \binom{10}{0} = 5 \times 1 = 5 \]
Total favorable ways for the numerator \( = 100 + 5 = 105 \).

Step 3: Calculating final probability.
\[ \text{Probability} = \frac{105}{1155} = \frac{1}{11} \] Let us re-verify base alignment for conditional variables. If the total space is normalized by raw permutations, option indicators align with index configurations. Let us mark (C) vs alternate target ratios.