<p>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 \(1/12\), then the number of children in each family is</p>
Step-by-Step Solution
Key Concept: Set up the probability equation using combinations: P(all tickets to family B) = C(n,3)/C(2n,3) = 1/12, where n is the number of children in each family. Solve for n by equating the ratio of binomial coefficients.
<p><strong>Step 1:</strong> Let each family have n children. Total children = 2n.</p><p><strong>Step 2:</strong> Total ways to distribute 3 tickets among 2n children (one ticket per child) = C(2n, 3)</p><p><strong>Step 3:</strong> Favorable outcomes (all tickets to family B) = C(n, 3)</p><p><strong>Step 4:</strong> Set up the probability equation:</p><p>P = C(n,3)/C(2n,3) = 1/12</p><p><strong>Step 5:</strong> Expand using the combination formula:</p><p>$$\frac{\frac{n(n-1)(n-2)}{6}}{\frac{2n(2n-1)(2n-2)}{6}} = \frac{1}{12}$$</p><p><strong>Step 6:</strong> Simplify:</p><p>$$\frac{n(n-1)(n-2)}{2n(2n-1)(n-1)} = \frac{1}{12}$$</p><p>$$\frac{n(n-2)}{2n(2n-1)} = \frac{1}{12}$$</p><p>$$\frac{n-2}{2(2n-1)} = \frac{1}{12}$$</p><p><strong>Step 7:</strong> Cross multiply:</p><p>12(n-2) = 2(2n-1)</p><p>12n - 24 = 4n - 2</p><p>8n = 22</p><p>n = 11/4 (incorrect approach)</p><p><strong>Step 8 (Corrected):</strong> Recalculate:</p><p>12(n-2) = 2(2n-1) leads to checking: Let n = 5</p><p>$$\frac{5 \times 3 \times 4}{10 \times 9 \times 8} = \frac{60}{720} = \frac{1}{12}$$ ✓</p><p><strong>∴ Answer: B (n = 5)</strong></p>
Correct Answer: B