<p>An unbiased dice is rolled until a number greater than 4 appears. What is the probability that an even number of tosses is needed?</p>
Step-by-Step Solution
Key Concept: This is an infinite series problem where we sum probabilities of getting the desired outcome (>4) on even-numbered tosses. Each toss before the final one must show ≤4, and the final toss must show >4 (which is always even).
<p><strong>Step 1:</strong> Identify outcomes. Numbers >4 are {5,6}, so P(>4) = 2/6 = 1/3. Numbers ≤4 are {1,2,3,4}, so P(≤4) = 4/6 = 2/3.</p><p><strong>Step 2:</strong> For even number of tosses, we need: (≤4, ≤4, ..., ≤4, >4) with exactly 2, 4, 6, ... tosses.</p><p><strong>Step 3:</strong> P(even tosses) = P(2 tosses) + P(4 tosses) + P(6 tosses) + ...</p><p>= (2/3)·(1/3) + (2/3)³·(1/3) + (2/3)⁵·(1/3) + ...</p><p>= (1/3)·(2/3)·[1 + (2/3)² + (2/3)⁴ + ...]</p><p><strong>Step 4:</strong> The bracketed part is a geometric series with first term a=1 and ratio r=(2/3)² = 4/9.</p><p>Sum = 1/(1 - 4/9) = 1/(5/9) = 9/5</p><p><strong>Step 5:</strong> P(even tosses) = (1/3)·(2/3)·(9/5) = (2/9)·(9/5) = 2/5</p><p>∴ Answer: <strong>2/5</strong></p>
Correct Answer: 2/5