Question

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 the children of the family $B$ is $\frac{1}{12}$, then the number of children in each family is?

• A. $4$
• B. $6$
• C. $3$
• D. $5$

Answer 1

The correct option is D $5$

There are two different families $A$ and $B$ with equal number of children.

Let the children in each family be $x$.

Thus the total number of children in both the families are $2x$

Now, it is given that $3$ tickets are distributed amongst the children of the families.

And all the tickets are distributed to the children in family $B$.

Thus, the probability that all the three tickets go to the children in family $B$ is given by

$\frac{1}{12}=\frac{{}^x{C}_3}{{}^{2x}{C}_3}$

On solving the above equation, we get,

Thus, $\frac{1}{12}=\frac{x(x-1)(x-2)}{2x(2x-1)(2x-2)}$

Thus, $\frac{1}{3}=\frac{x-2}{2x-1}$

$\to 3x-6=2x-1$

$\to x=5$

Thus, the number of children in each family are $5$.