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 : (a) 3 (b) 5 (c) 4 (d) 6

Options

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

Answer

To solve the problem step by step, let's denote the number of children in each family as \( x \). Therefore, the total number of children from both families A and B is \( 2x \).

### Step 1: Determine the total ways to distribute tickets
The total number of ways to distribute 3 tickets among \( 2x \) children is given by the combination formula:

\[ \text{Total ways} = \binom{2x}{3} = \frac{2x(2x-1)(2x-2)}{3!} = \frac{2x(2x-1)(2x-2)}{6} \]

### Step 2: Determine the ways to distribute tickets to Family B
The number of ways to distribute all 3 tickets to the children of Family B (which has \( x \) children) is:

\[ \text{Ways to Family B} = \binom{x}{3} = \frac{x(x-1)(x-2)}{3!} = \frac{x(x-1)(x-2)}{6} \]

### Step 3: Set up the probability equation
According to the problem, the probability that all tickets go to Family B is given as \( \frac{1}{12} \). Therefore, we can set up the equation:

\[ \text{Probability} = \frac{\text{Ways to Family B}}{\text{Total ways}} = \frac{\binom{x}{3}}{\binom{2x}{3}} = \frac{\frac{x(x-1)(x-2)}{6}}{\frac{2x(2x-1)(2x-2)}{6}} = \frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)} \]

Setting this equal to \( \frac{1}{12} \):

\[ \frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)} = \frac{1}{12} \]

### Step 4: Simplify the equation
Cross-multiplying gives:

\[ 12 \cdot x(x-1)(x-2) = 2x(2x-1)(2x-2) \]

Dividing both sides by \( x \) (assuming \( x \neq 0 \)):

\[ 12(x-1)(x-2) = 2(2x-1)(2x-2) \]

### Step 5: Expand both sides
Expanding both sides:

Left side:

\[ 12(x^2 - 3x + 2) = 12x^2 - 36x + 24 \]

Right side:

\[ 2(4x^2 - 6x + 2) = 8x^2 - 12x + 4 \]

### Step 6: Set the equation to zero
Setting both sides equal gives:

\[ 12x^2 - 36x + 24 = 8x^2 - 12x + 4 \]

Rearranging terms:

\[ 12x^2 - 8x^2 - 36x + 12x + 24 - 4 = 0 \]

This simplifies to:

\[ 4x^2 - 24x + 20 = 0 \]

### Step 7: Simplify the quadratic equation
Dividing the entire equation by 4:

\[ x^2 - 6x + 5 = 0 \]

### Step 8: Factor the quadratic
Factoring gives:

\[ (x - 1)(x - 5) = 0 \]

Thus, \( x = 1 \) or \( x = 5 \).

### Step 9: Determine the valid solution
Since the problem states that both families have children, we discard \( x = 1 \) (as it does not fit the context of the problem). Therefore, the number of children in each family is:

\[ \boxed{5} \]