<p>Let \(P(A)\) be the probability of at least two girls, and \(P(B)\) be the probability of all girls in a family with 4 children (probability of a girl = \(\dfrac{1}{4}\)). Find the required conditional probability \(\dfrac{P(B \cap A)}{P(A)}\).</p>
Step-by-Step Solution
Key Concept: Recognize that event B (all girls) is a subset of event A (at least two girls), so B ∩ A = B. The conditional probability simplifies to P(B)/P(A), requiring careful calculation of both probabilities.
<p><strong>Step 1: Identify the relationship between events</strong></p><p>Event B = {all 4 children are girls} is a subset of Event A = {at least 2 girls}. Therefore, B ∩ A = B.</p><p><strong>Step 2: Calculate P(A) - Probability of at least 2 girls</strong></p><p>P(at least 2 girls) = 1 - P(0 girls) - P(1 girl)</p><p>P(0 girls) = C(4,0)(1/4)⁰(3/4)⁴ = (81/256)</p><p>P(1 girl) = C(4,1)(1/4)¹(3/4)³ = 4 · (1/4) · (27/64) = (108/256)</p><p>P(A) = 1 - 81/256 - 108/256 = 67/256</p><p><strong>Step 3: Calculate P(B) - Probability of all 4 girls</strong></p><p>P(B) = (1/4)⁴ = 1/256</p><p><strong>Step 4: Apply conditional probability formula</strong></p><p>P(B|A) = P(B ∩ A)/P(A) = P(B)/P(A) = (1/256)/(67/256) = 1/67</p><p>∴ Answer: A (which equals 1/67)</p>
Correct Answer: A