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

Step 1: Understand the condition.
We want the chance of having more women than men, given the committee already has at least one woman. So both top and bottom counts must respect at least one woman.
Step 2: Count the conditioned total.
All 4-member groups: $\binom{15}{4} = 1365$. Groups with no women: $\binom{10}{4} = 210$. So with at least one woman: $1365 - 210 = 1155$.
Step 3: Count more women than men, case 1.
3 women and 1 man: $\binom{5}{3}\binom{10}{1} = 10\times10 = 100$.
Step 4: Case 2.
4 women and 0 men: $\binom{5}{4}\binom{10}{0} = 5\times1 = 5$.
Step 5: Add the favorable.
Favorable $= 100 + 5 = 105$.
Step 6: Form the probability.
\[ P = \frac{105}{1155} = \frac{1}{11} \] \[ \boxed{\tfrac{1}{11}} \]