Question:
Two different families A and B are blessed with equal numbe 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 :
Two different families A and B are blessed with equal numbe 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 :
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With n children per family, the probability is \(\binom{n}{3}/\binom{2n}{3}\); set it equal to 1/12 and solve for n.
Updated On: Nov 11, 2026
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The Correct Option is C
Solution and Explanation
Let each family have n children (total 2n). Tickets go to 3 distinct children.
\(P = \frac{\binom{n}{3}}{\binom{2n}{3}} = \frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} = \frac{n-2}{4(2n-1)}\).
Setting \(\frac{n-2}{4(2n-1)} = \frac{1}{12}\): \(3(n-2) = 2n-1\), so \(n = 5\).
Check: \(\binom{5}{3}/\binom{10}{3} = \frac{10}{120} = \frac{1}{12}\).
Answer: (3) 5
\(P = \frac{\binom{n}{3}}{\binom{2n}{3}} = \frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)} = \frac{n-2}{4(2n-1)}\).
Setting \(\frac{n-2}{4(2n-1)} = \frac{1}{12}\): \(3(n-2) = 2n-1\), so \(n = 5\).
Check: \(\binom{5}{3}/\binom{10}{3} = \frac{10}{120} = \frac{1}{12}\).
Answer: (3) 5
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