<p>A six-faced dice is so biased that it is twice as likely to show an even number as an odd number when thrown. It is thrown twice, the probability that the sum of two numbers thrown is even is</p>
Step-by-Step Solution
Key Concept: The sum of two numbers is even when both are even or both are odd. Use the given bias condition to find individual probabilities, then apply the addition rule for mutually exclusive events.
<p><strong>Step 1:</strong> Find probability of even and odd for a single throw.</p><p>Let P(odd) = p, then P(even) = 2p (given: even is twice as likely)</p><p>Since p + 2p = 1, we get 3p = 1, so p = 1/3</p><p>Therefore: P(odd) = 1/3 and P(even) = 2/3</p><p><strong>Step 2:</strong> Find probability that sum of two throws is even.</p><p>Sum is even when: (both even) OR (both odd)</p><p>P(sum is even) = P(both even) + P(both odd)</p><p>= P(even) × P(even) + P(odd) × P(odd)</p><p>= (2/3)² + (1/3)²</p><p>= 4/9 + 1/9</p><p>= 5/9</p><p><strong>Step 3:</strong> Verify the result.</p><p>P(sum is odd) = 2 × P(even) × P(odd) = 2 × (2/3) × (1/3) = 4/9</p><p>Check: 5/9 + 4/9 = 1 ✓</p><p>∴ Answer: <strong>5/9</strong> (Option D)</p>
Correct Answer: D