<p>A fair die is tossed repeatedly. \(A\) wins if it is 1 or 2 on two consecutive tosses and \(B\) wins if it is 3, 4, 5 or 6 on two consecutive tosses. The probability that \(A\) wins if the die is tossed indefinitely is</p>
Step-by-Step Solution
Key Concept: Model this as a Markov chain with states based on the last toss outcome. Set up equations where P(A wins | last toss was i) accounts for immediate win conditions and transitions to other states.
<p><strong>Step 1:</strong> Define states based on last toss. Let p<sub>i</sub> = probability A wins given last toss was i.</p><p><strong>Step 2:</strong> For A to win: need two consecutive rolls from {1,2}. P(roll 1 or 2) = 1/3, P(roll 3,4,5,6) = 2/3.</p><p><strong>Step 3:</strong> If last toss was 1 or 2 (A's winning numbers):<br/>• Next roll is 1 or 2 with probability 1/3 → A wins<br/>• Next roll is 3,4,5,6 with probability 2/3 → B wins<br/>So p₁ = p₂ = 1/3</p><p><strong>Step 4:</strong> If last toss was 3,4,5,6 (B's numbers):<br/>• Next roll is 1 or 2 with probability 1/3 → go to state 1 or 2<br/>• Next roll is 3,4,5,6 with probability 2/3 → B wins<br/>So p₃ = p₄ = p₅ = p₆ = (1/3)·(1/3) = 1/9</p><p><strong>Step 5:</strong> Initially, first toss outcome is equally likely. Overall probability A wins:<br/>P(A wins) = (1/6)(p₁ + p₂ + p₃ + p₄ + p₅ + p₆)<br/>= (1/6)(1/3 + 1/3 + 1/9 + 1/9 + 1/9 + 1/9)<br/>= (1/6)(2/3 + 4/9)<br/>= (1/6)(6/9 + 4/9)<br/>= (1/6)(10/9)<br/>= 10/54 = <strong>5/27</strong></p><p>∴ Answer: B</p>
Correct Answer: B