A committee of 5 persons is to be randomly selected from a group of 5 men and 4 women and a chairperson will be randomly selected from the committee will have exactly 2 women and 3 men and the chairperson will be a man is p, then `(1)/(p)` is equal to

Answer

To solve the problem step by step, we need to calculate the probability \( p \) that a randomly selected committee of 5 persons, consisting of exactly 2 women and 3 men, has a chairperson who is a man. We will then find \( \frac{1}{p} \).

### Step 1: Calculate the total number of ways to select 5 persons from 9 (5 men and 4 women)

The total number of ways to select 5 persons from a group of 9 (5 men + 4 women) is given by the combination formula \( \binom{n}{r} \):

\[ \text{Total ways} = \binom{9}{5} \]

### Step 2: Calculate the number of ways to select 2 women and 3 men

Next, we need to calculate the number of ways to select 2 women from 4 and 3 men from 5:

\[ \text{Ways to select 2 women} = \binom{4}{2} \]
\[ \text{Ways to select 3 men} = \binom{5}{3} \]

Thus, the total number of ways to select 2 women and 3 men is:

\[ \text{Ways to select 2 women and 3 men} = \binom{4}{2} \times \binom{5}{3} \]

### Step 3: Calculate the probability \( p \)

The probability \( p \) that a committee has exactly 2 women and 3 men is given by:

\[ p = \frac{\text{Ways to select 2 women and 3 men}}{\text{Total ways}} \]

Substituting the values we calculated:

\[ p = \frac{\binom{4}{2} \times \binom{5}{3}}{\binom{9}{5}} \]

### Step 4: Calculate the probability that the chairperson is a man

In the committee of 5 persons, there are 3 men. The probability that the chairperson selected is a man is:

\[ \text{Probability chairperson is a man} = \frac{3}{5} \]

### Step 5: Combine the probabilities

The total probability \( P \) that a committee has exactly 2 women and 3 men, and the chairperson is a man is:

\[ P = p \times \frac{3}{5} \]

### Step 6: Calculate \( \frac{1}{P} \)

Finally, we need to find \( \frac{1}{P} \):

\[ \frac{1}{P} = \frac{1}{p \times \frac{3}{5}} = \frac{5}{3p} \]

### Step 7: Substitute \( p \) into \( \frac{1}{P} \)

Now we substitute \( p \) into the equation:

\[ \frac{1}{P} = \frac{5}{3 \left( \frac{\binom{4}{2} \times \binom{5}{3}}{\binom{9}{5}} \right)} = \frac{5 \cdot \binom{9}{5}}{3 \cdot \binom{4}{2} \cdot \binom{5}{3}} \]

### Step 8: Calculate the values of combinations

Calculating the combinations:

- \( \binom{4}{2} = 6 \)
- \( \binom{5}{3} = 10 \)
- \( \binom{9}{5} = 126 \)

Substituting these values:

\[ \frac{1}{P} = \frac{5 \cdot 126}{3 \cdot 6 \cdot 10} = \frac{630}{180} = \frac{21}{6} = 3.5 \]

### Final Answer

Thus, the final answer is:

\[ \frac{1}{p} = 3.5 \]