From a group of 8 boys and 3 girls, a committee of 5 members to be formed. Find the probability that 2 particular girls are included in the committee.

Options

A. `(4)/(11)`
B. `(2)/(11)`
C. `(6)/(11)`
D. `(8)/(11)`

Explanation

To solve the problem of finding the probability that 2 particular girls are included in a committee of 5 members formed from a group of 8 boys and 3 girls, we can follow these steps:

### Step 1: Calculate the total number of ways to form a committee of 5 members from 11 people.
We have a total of 11 people (8 boys + 3 girls). The total number of ways to choose 5 members from 11 is given by the combination formula:

\[ \text{Total ways} = \binom{11}{5} \]

Calculating this:

\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

### Step 2: Calculate the number of ways to form a committee including the 2 particular girls.
If we want to include 2 particular girls in the committee, we have already chosen 2 members. Therefore, we need to choose 3 more members from the remaining people.

After including the 2 girls, we have:
- 1 girl left (since we started with 3 girls and included 2)
- 8 boys

Thus, we have a total of 9 people remaining (8 boys + 1 girl). We need to choose 3 members from these 9:

\[ \text{Ways to choose 3 from 9} = \binom{9}{3} \]

Calculating this:

\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

### Step 3: Calculate the probability.
The probability that the committee includes the 2 particular girls is given by the ratio of the number of favorable outcomes to the total outcomes:

\[ \text{Probability} = \frac{\text{Ways to choose committee with 2 girls}}{\text{Total ways to choose committee}} = \frac{84}{462} \]

Simplifying this fraction:

\[ \frac{84}{462} = \frac{2}{11} \]

### Final Answer:
The probability that the committee includes the 2 particular girls is \(\frac{2}{11}\).

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