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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?
$4$
$6$
$3$
$5$
Solution
Let the number of children ineach familybe $x$.
Thus the total number of children in both the families are $2 x$
Now, it is given that $3$ tickets are distributed amongst the children of these two families.
Thus, the probability that all the three tickets go to the children in family $B$
$=\frac{^{x} C_{3}}{^{2 x} C_{3}}=\frac{1}{12}$
$\Rightarrow \frac{x(x-1)(x-2)}{2 x(2 x-1)(2 x-2)}=\frac{1}{12}$
$\Rightarrow \frac{(x-2)}{(2 x-1)}=\frac{1}{6}$
$\Rightarrow x=5$
Thus, the number of children in each family is $5$.
