There are 3 men, 2 women and 4 children. If 4 persons are selected then which of the following is/are true?

Options

A. Probability of selecting exactly 2 children = `10/21`
B. Probability of selecting atmost 2 children = `5/6`
C. Probability of selecting exactly 2 children = `5/21`
D. Probability of selecting atmost 2 children = `25/42`

Answer

To solve the problem step by step, we need to find the probabilities of selecting exactly 2 children and at most 2 children from a group of 3 men, 2 women, and 4 children when selecting 4 persons in total.

### Step 1: Determine the total number of people
We have:
- 3 men
- 2 women
- 4 children

Total number of people = 3 + 2 + 4 = 9

### Step 2: Calculate the total ways to select 4 persons from 9
The total number of ways to choose 4 persons from 9 is given by the combination formula:
\[ \text{Total ways} = \binom{9}{4} \]
Calculating this:
\[ \binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

### Step 3: Calculate the probability of selecting exactly 2 children
To select exactly 2 children, we need to select:
- 2 children from 4
- 2 persons from the remaining 5 (3 men + 2 women)

The number of ways to do this is:
\[ \text{Ways to select 2 children} = \binom{4}{2} \]
\[ \text{Ways to select 2 from 5 (men and women)} = \binom{5}{2} \]
Thus, the total ways to select exactly 2 children and 2 others:
\[ \text{Total ways for exactly 2 children} = \binom{4}{2} \times \binom{5}{2} \]

Calculating:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Thus:
\[ \text{Total ways for exactly 2 children} = 6 \times 10 = 60 \]

### Step 4: Calculate the probability of selecting exactly 2 children
The probability of selecting exactly 2 children is:
\[ P(\text{exactly 2 children}) = \frac{\text{Ways to select exactly 2 children}}{\text{Total ways to select 4 persons}} = \frac{60}{126} \]
Simplifying:
\[ P(\text{exactly 2 children}) = \frac{10}{21} \]

### Step 5: Calculate the probability of selecting at most 2 children
To find the probability of selecting at most 2 children, we consider three cases: selecting 0, 1, or 2 children.

#### Case 1: Selecting 0 children
\[ \text{Ways} = \binom{4}{0} \times \binom{5}{4} = 1 \times 5 = 5 \]

#### Case 2: Selecting 1 child
\[ \text{Ways} = \binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40 \]

#### Case 3: Selecting 2 children (already calculated)
\[ \text{Ways} = 60 \]

#### Total ways for at most 2 children
\[ \text{Total ways for at most 2 children} = 5 + 40 + 60 = 105 \]

### Step 6: Calculate the probability of selecting at most 2 children
The probability of selecting at most 2 children is:
\[ P(\text{at most 2 children}) = \frac{105}{126} \]
Simplifying:
\[ P(\text{at most 2 children}) = \frac{5}{6} \]

### Conclusion
1. The probability of selecting exactly 2 children is \( \frac{10}{21} \).
2. The probability of selecting at most 2 children is \( \frac{5}{6} \).