<p><strong>816.</strong> Consider a family of \(n\) children. Let two events \(A\) and \(B\) are defined as follows:<br>\(A\): is the event that the family has both boys and girls<br>\(B\): is the event that the family has atmost one girl<br>If the events \(A\) and \(B\) are independent, then find the value of \(n\).<br>[<strong>Note:</strong> Probability that a randomly selected child is a boy or girl is same.]</p>
Step-by-Step Solution
Key Concept: For independence of events A and B, we need P(A∩B) = P(A)·P(B). Event A∩B represents 'both boys and girls AND at most one girl' which means exactly one girl. Use this to set up an equation and solve for n.
<p><strong>Step 1: Identify the events clearly.</strong></p><p>A: Family has both boys and girls, so P(A) = 1 - P(all boys) - P(all girls) = 1 - 1/2^n - 1/2^n = 1 - 2/2^n = (2^n - 2)/2^n</p><p>B: At most one girl means 0 girls or 1 girl, so P(B) = P(0 girls) + P(1 girl) = 1/2^n + C(n,1)/2^n = (1 + n)/2^n</p><p><strong>Step 2: Find A∩B.</strong></p><p>A∩B is 'both boys and girls' AND 'at most one girl' = exactly one girl</p><p>P(A∩B) = C(n,1)/2^n = n/2^n</p><p><strong>Step 3: Apply independence condition P(A∩B) = P(A)·P(B).</strong></p><p>n/2^n = [(2^n - 2)/2^n] · [(1 + n)/2^n]</p><p>n/2^n = (2^n - 2)(1 + n)/2^(2n)</p><p>n · 2^n = (2^n - 2)(1 + n)</p><p>n · 2^n = 2^n + n·2^n - 2 - 2n</p><p>0 = 2^n - 2 - 2n</p><p>2^n = 2n + 2</p><p><strong>Step 4: Solve for n by testing values.</strong></p><p>n = 1: 2^1 = 2, but 2(1) + 2 = 4 ✗</p><p>n = 2: 2^2 = 4, and 2(2) + 2 = 6 ✗</p><p>n = 3: 2^3 = 8, and 2(3) + 2 = 8 ✓</p><p>∴ Answer: <strong>n = 3</strong></p>
Correct Answer: 0