A team of 6 persons is to be selected from a group of 3 men, 3 women and 4 children. What is the probability that it comprises of exactly 2 children?

Options

A. `1/35`
B. `1/10`
C. `3/70`
D. `3/7`

Answer

To solve the problem of selecting a team of 6 persons from a group of 3 men, 3 women, and 4 children, where we want exactly 2 children in the team, we can break down the solution into clear steps:

### Step-by-Step Solution:

1. **Identify Total Persons**:
- We have 3 men, 3 women, and 4 children.
- Total persons = 3 men + 3 women + 4 children = 10 persons.

2. **Select 2 Children**:
- We need to select exactly 2 children from the 4 available.
- The number of ways to choose 2 children from 4 is given by the combination formula \( \binom{n}{r} \), which is \( \binom{4}{2} \).

\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

3. **Select Remaining 4 Persons**:
- After selecting 2 children, we need to select 4 more persons from the remaining group of 8 persons (3 men + 3 women + 2 remaining children).
- We can have different combinations of men and women in these 4 persons. The possible distributions are:
- 3 men and 1 woman
- 2 men and 2 women
- 1 man and 3 women

4. **Calculate Each Case**:
- **Case 1**: 3 men and 1 woman
- Ways to choose 3 men from 3: \( \binom{3}{3} = 1 \)
- Ways to choose 1 woman from 3: \( \binom{3}{1} = 3 \)
- Total ways for this case: \( 6 \times 1 \times 3 = 18 \)

- **Case 2**: 2 men and 2 women
- Ways to choose 2 men from 3: \( \binom{3}{2} = 3 \)
- Ways to choose 2 women from 3: \( \binom{3}{2} = 3 \)
- Total ways for this case: \( 6 \times 3 \times 3 = 54 \)

- **Case 3**: 1 man and 3 women
- Ways to choose 1 man from 3: \( \binom{3}{1} = 3 \)
- Ways to choose 3 women from 3: \( \binom{3}{3} = 1 \)
- Total ways for this case: \( 6 \times 3 \times 1 = 18 \)

5. **Total Ways to Select Team with Exactly 2 Children**:
- Adding all cases together:
\[ 18 + 54 + 18 = 90 \]

6. **Total Ways to Select Any 6 Persons from 10**:
- The total number of ways to select any 6 persons from 10 is given by \( \binom{10}{6} \):
\[ \binom{10}{6} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

7. **Calculate Probability**:
- The probability of selecting a team of 6 persons that includes exactly 2 children is given by the ratio of the favorable outcomes to the total outcomes:
\[ P(\text{exactly 2 children}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{90}{210} = \frac{3}{7} \]

### Final Answer:
The probability that the team comprises exactly 2 children is \( \frac{3}{7} \).