Question

A committee of two persons is selected from two men and two women. What is the probability that the committee will have one man?

• A. \(\frac{1}{3}\)
• B. \(\frac{2}{3}\)
• C. \(1\)
• D. \(\frac{4}{3}\)

Hint:

For committee-selection problems, use combinations because the order of selection is irrelevant. Always count favourable committees and total committees separately before computing probability.

Answer 1

The Correct Option is B

Concept: When selecting a committee, order does not matter. Therefore, combinations are used. The probability is \[ P(E) = \frac{\text{Number of favourable committees}} {\text{Total number of committees}}. \]

Step 1: Find the total number of committees. There are \[ 2+2=4 \] people. The number of ways to choose a committee of two persons is \[ {}^{4}C_{2} = \frac{4!}{2!2!} = 6. \] Thus, \[ n(S)=6. \]

Step 2: Find the number of favourable committees. The committee must contain exactly one man. Choose \[ 1 \] man from \[ 2 \] men: \[ {}^{2}C_{1}=2. \] Choose \[ 1 \] woman from \[ 2 \] women: \[ {}^{2}C_{1}=2. \] Therefore, \[ n(E) = {}^{2}C_{1}\times{}^{2}C_{1} = 2\times2 = 4. \]

Step 3: Compute the probability. \[ P(E) = \frac{4}{6} = \frac{2}{3}. \] Conclusion: Hence, the probability that the committee contains exactly one man is \[ \boxed{\frac{2}{3}}. \]