<p>A coin is tossed three times. Let \(A\): at least two heads, \(B\): at most two heads. Find \(P(A/B)\).</p>
Step-by-Step Solution
Key Concept: Conditional probability P(A/B) = P(A∩B)/P(B). Recognize that A∩B contains outcomes with exactly two heads (the intersection of 'at least 2' and 'at most 2'), while B includes all outcomes except HHH.
<p><strong>Step 1:</strong> List all outcomes and identify sets A and B.</p><p>Sample space: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8</p><p><strong>Step 2:</strong> Find set A (at least two heads).</p><p>A = {HHH, HHT, HTH, THH}, n(A) = 4</p><p><strong>Step 3:</strong> Find set B (at most two heads).</p><p>B = {HHT, HTH, HTT, THH, THT, TTH, TTT}, n(B) = 7</p><p><strong>Step 4:</strong> Find A∩B (at least two AND at most two = exactly two heads).</p><p>A∩B = {HHT, HTH, THH}, n(A∩B) = 3</p><p><strong>Step 5:</strong> Apply conditional probability formula.</p><p>P(A/B) = P(A∩B)/P(B) = [n(A∩B)/n(S)] / [n(B)/n(S)] = n(A∩B)/n(B) = 3/7</p><p>∴ Answer: B</p>
Correct Answer: B