Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to children of the family B is `(1)/(12)` then the number of children in each family is :

Options

A. 3
B. 5
C. 4
D. 6

Answer

To solve the problem, we need to find the number of children in each family given the probability that all tickets go to children of family B is \( \frac{1}{12} \).

### Step-by-Step Solution:

1. **Define Variables:**
Let \( x \) be the number of children in each family (A and B). Thus, the total number of children is \( 2x \).

2. **Total Ways to Distribute Tickets:**
The total number of ways to distribute 3 tickets among \( 2x \) children is given by the combination formula:
\[ \text{Total ways} = \binom{2x}{3} = \frac{(2x)!}{3!(2x-3)!} \]

3. **Ways to Give All Tickets to Family B:**
The number of ways to distribute all 3 tickets to the children of family B (which has \( x \) children) is:
\[ \text{Ways to give tickets to B} = \binom{x}{3} = \frac{x!}{3!(x-3)!} \]

4. **Probability Calculation:**
The probability that all tickets go to family B is given by:
\[ P(\text{All to B}) = \frac{\text{Ways to give tickets to B}}{\text{Total ways}} = \frac{\binom{x}{3}}{\binom{2x}{3}} \]
According to the problem, this probability is \( \frac{1}{12} \):
\[ \frac{\binom{x}{3}}{\binom{2x}{3}} = \frac{1}{12} \]

5. **Cross Multiplying:**
Cross-multiplying gives:
\[ 12 \cdot \binom{x}{3} = \binom{2x}{3} \]

6. **Expanding Combinations:**
Expanding both sides using the combination formula:
\[ 12 \cdot \frac{x(x-1)(x-2)}{6} = \frac{(2x)(2x-1)(2x-2)}{6} \]
Simplifying gives:
\[ 2x(x-1)(x-2) = (2x)(2x-1)(2x-2) \]

7. **Cancelling Common Factors:**
We can cancel \( 2x \) from both sides (assuming \( x \neq 0 \)):
\[ (x-1)(x-2) = (2x-1)(2x-2) \]

8. **Expanding Both Sides:**
Expanding both sides:
\[ x^2 - 3x + 2 = 4x^2 - 6x + 2 \]

9. **Rearranging the Equation:**
Rearranging gives:
\[ 0 = 4x^2 - 6x + 2 - x^2 + 3x - 2 \]
Simplifying results in:
\[ 0 = 3x^2 - 3x \]

10. **Factoring the Quadratic:**
Factoring out \( 3x \):
\[ 3x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

11. **Finding Valid Solutions:**
Since \( x \) represents the number of children, it must be a positive integer. Therefore, we discard \( x = 0 \) and consider \( x = 5 \).

### Conclusion:
The number of children in each family is \( x = 5 \).