A committee of three has to be chosen from a group of 4 men and 5 women. If the selection is made at random, what is the probability that exactly two members are men?

Options

A. `5//14`
B. `1//21`
C. `3//14`
D. `8//21`

Answer

To solve the problem of finding the probability that exactly two members of a committee of three chosen from a group of 4 men and 5 women are men, we can follow these steps:

### Step 1: Determine the total number of ways to choose 3 members from 9 people.
We have a total of 4 men and 5 women, which gives us a total of 9 people. The number of ways to choose 3 members from 9 can be calculated using the combination formula:

\[ \text{Total ways} = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \]

### Step 2: Calculate \(\binom{9}{3}\).
Calculating this gives:

\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Step 3: Determine the number of ways to choose exactly 2 men and 1 woman.
To have exactly 2 men in the committee, we need to choose 2 men from the 4 available men and 1 woman from the 5 available women. The number of ways to choose 2 men from 4 is:

\[ \text{Ways to choose 2 men} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

The number of ways to choose 1 woman from 5 is:

\[ \text{Ways to choose 1 woman} = \binom{5}{1} = 5 \]

### Step 4: Calculate the total number of favorable outcomes.
The total number of ways to choose 2 men and 1 woman is the product of the two combinations calculated above:

\[ \text{Favorable outcomes} = \binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30 \]

### Step 5: Calculate the probability.
The probability of choosing exactly 2 men and 1 woman is given by the ratio of the number of favorable outcomes to the total number of outcomes:

\[ P(\text{exactly 2 men}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{30}{84} \]

### Step 6: Simplify the probability.
To simplify \(\frac{30}{84}\):

\[ P(\text{exactly 2 men}) = \frac{30 \div 6}{84 \div 6} = \frac{5}{14} \]

Thus, the probability that exactly two members are men is:

\[ \boxed{\frac{5}{14}} \]

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