<p>Three coins are tossed. If one of them shows tail, then find the probability that all three coins show tail.</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(all tails | at least one tail) = P(all tails AND at least one tail) / P(at least one tail). Since 'all tails' automatically satisfies 'at least one tail', the numerator simplifies to P(all tails).
<p><strong>Step 1:</strong> Total outcomes when tossing 3 coins = 2³ = 8</p><p><strong>Step 2:</strong> Let A = event that all three coins show tail = {TTT}</p><p>P(A) = 1/8</p><p><strong>Step 3:</strong> Let B = event that at least one coin shows tail</p><p>Complement: P(no tail) = P(HHH) = 1/8</p><p>So P(B) = 1 - 1/8 = 7/8</p><p><strong>Step 4:</strong> We need P(A|B) = P(A ∩ B) / P(B)</p><p>Since all tails includes at least one tail: A ∩ B = A</p><p>P(A|B) = (1/8) / (7/8) = 1/8 × 8/7 = 1/7</p><p>∴ <strong>Answer: 1/7</strong></p>
Correct Answer: 1/7