A commitree of 5 principals is to be selected from a group of 6 gent principals and 8 lady principals. If the selection is made randomly , find the probability that there are 3 lady principals and 2 gent principals .

Answer

To solve the problem of selecting a committee of 5 principals from a group of 6 gent principals and 8 lady principals, where we want to find the probability of selecting 3 lady principals and 2 gent principals, we can follow these steps:

### Step 1: Determine the total number of principals
We have:
- 6 gent principals
- 8 lady principals

Total principals = 6 + 8 = 14

### Step 2: Calculate the total number of ways to select 5 principals from 14
The total number of ways to choose 5 principals from 14 is given by the combination formula:
\[ \text{Total outcomes} = \binom{14}{5} \]

### Step 3: Calculate the number of favorable outcomes
We want to select:
- 3 lady principals from 8
- 2 gent principals from 6

The number of ways to choose 3 lady principals from 8 is:
\[ \text{Ways to choose 3 ladies} = \binom{8}{3} \]

The number of ways to choose 2 gent principals from 6 is:
\[ \text{Ways to choose 2 gents} = \binom{6}{2} \]

### Step 4: Calculate the total number of favorable outcomes
The total number of favorable outcomes (selecting 3 lady principals and 2 gent principals) is the product of the two combinations calculated above:
\[ \text{Favorable outcomes} = \binom{8}{3} \times \binom{6}{2} \]

### Step 5: Calculate the probability
The probability \( P \) of selecting 3 lady principals and 2 gent principals is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{\binom{8}{3} \times \binom{6}{2}}{\binom{14}{5}} \]

### Step 6: Compute the combinations
Now we compute the combinations:
1. \(\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
2. \(\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15\)
3. \(\binom{14}{5} = \frac{14!}{5!(14-5)!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002\)

### Step 7: Substitute the values into the probability formula
Now substituting the values we calculated:
\[ P = \frac{56 \times 15}{2002} = \frac{840}{2002} \]

### Step 8: Simplify the fraction
To simplify \(\frac{840}{2002}\), we can divide both the numerator and denominator by their greatest common divisor (GCD), which is 2:
\[ P = \frac{420}{1001} \]

### Final Answer
Thus, the probability that there are 3 lady principals and 2 gent principals in the committee is:
\[ P = \frac{420}{1001} \]