<p>A coin is tossed three times. Let \(A\): at most two tails, \(B\): at least one tail. Find \(P(A/B)\).</p>
Step-by-Step Solution
Key Concept: Use the conditional probability formula P(A/B) = P(A∩B)/P(B). Recognize that A∩B consists of outcomes with at least one tail but at most two tails (excluding all three tails and zero tails).
<p><strong>Step 1:</strong> List all outcomes when coin is tossed 3 times: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Total = 8 outcomes.</p><p><strong>Step 2:</strong> Find event B (at least one tail): {HHT, HTH, HTT, THH, THT, TTH, TTT}. So |B| = 7, and P(B) = 7/8.</p><p><strong>Step 3:</strong> Find event A (at most two tails): {HHH, HHT, HTH, HTT, THH, THT, TTH}. This excludes TTT. So |A| = 7.</p><p><strong>Step 4:</strong> Find A∩B (at least one tail AND at most two tails): {HHT, HTH, HTT, THH, THT, TTH}. This is all of B except TTT. So |A∩B| = 6, and P(A∩B) = 6/8.</p><p><strong>Step 5:</strong> Apply conditional probability formula: P(A/B) = P(A∩B)/P(B) = (6/8)/(7/8) = 6/7.</p><p>∴ Answer: B</p>
Correct Answer: B