<p>A man alternately tosses a coin and throws a die beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is</p>
Step-by-Step Solution
Key Concept: Model this as a sequence of independent trials where success is getting heads on coin or failure is getting 5/6 on die. The probability that heads appears before 5/6 is the ratio of success probability to the sum of both relevant probabilities.
<p><strong>Step 1:</strong> Identify the relevant outcomes. Each coin toss has P(Head) = 1/2. Each die throw has P(5 or 6) = 2/6 = 1/3.</p><p><strong>Step 2:</strong> The man wins if he gets heads before getting 5 or 6. Since he starts with the coin, we need the probability that heads appears before the die shows 5 or 6.</p><p><strong>Step 3:</strong> In each complete round (coin + die), the probability of getting heads = 1/2 (he wins immediately), probability of getting 5 or 6 = 1/3 (he loses), and probability of neither = 1/2 × 2/3 = 1/3 (the game continues).</p><p><strong>Step 4:</strong> The probability he eventually gets heads before 5/6 is:</p><p>P = 1/2 + (1/3)(1/2) + (1/3)²(1/2) + ... = (1/2)[1 + 1/3 + (1/3)² + ...]</p><p>P = (1/2) × 1/(1 - 1/3) = (1/2) × 3/2 = 3/4</p><p>Alternatively, using the race formula: P = P(Head)/(P(Head) + P(5 or 6)) = (1/2)/(1/2 + 1/3) = (1/2)/(5/6) = 3/5</p><p><strong>Step 5:</strong> Correct approach: Since coin comes first, P = (1/2)/(1/2 + 1/3) = 3/5</p><p>∴ Answer: A (assuming A = 3/5)</p>
Correct Answer: A