<p>Urn <em>A</em> contains 6 red and 4 black balls and urn <em>B</em> contains 4 red and 6 black balls. One ball is drawn at random from urn <em>A</em> and placed in urn <em>B</em>. Then one ball is drawn at random from urn <em>B</em> and placed in urn <em>A</em>. If one ball is now drawn at random from urn <em>A</em>, the probability that it is red is:</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by tracking all possible states of urn A after the two transfers. The final composition of urn A depends on whether a red or black ball was transferred in each step.
<p><strong>Step 1: Initial Setup</strong></p><p>Urn A: 6 red, 4 black (total 10)</p><p>Urn B: 4 red, 6 black (total 10)</p><p><strong>Step 2: First Transfer (A → B)</strong></p><p>P(red from A) = 6/10 = 3/5</p><p>P(black from A) = 4/10 = 2/5</p><p><strong>Step 3: Second Transfer (B → A) - Case Analysis</strong></p><p><strong>Case 1:</strong> Red transferred from A to B (prob = 3/5)</p><p>After transfer: Urn B has 5 red, 6 black</p><p>P(red from B) = 5/11 → Urn A becomes: 6 red, 4 black</p><p>P(black from B) = 6/11 → Urn A becomes: 5 red, 5 black</p><p><strong>Case 2:</strong> Black transferred from A to B (prob = 2/5)</p><p>After transfer: Urn B has 4 red, 7 black</p><p>P(red from B) = 4/11 → Urn A becomes: 7 red, 3 black</p><p>P(black from B) = 7/11 → Urn A becomes: 6 red, 4 black</p><p><strong>Step 4: Final Probability</strong></p><p>P(red from final A) = P(Case 1, red) × P(red|6R,4B) + P(Case 1, black) × P(red|5R,5B) + P(Case 2, red) × P(red|7R,3B) + P(Case 2, black) × P(red|6R,4B)</p><p>= (3/5)(5/11)(6/10) + (3/5)(6/11)(5/10) + (2/5)(4/11)(7/10) + (2/5)(7/11)(6/10)</p><p>= (3/5)(1/11)[5×6/10 + 6×5/10] + (2/5)(1/11)[4×7/10 + 7×6/10]</p><p>= (3/5)(1/11)(60/10) + (2/5)(1/11)(70/10)</p><p>= (3/5)(1/11)(6) + (2/5)(1/11)(7)</p><p>= (1/55)[18/5 + 14/5] = (1/55)(32/5) = 32/55</p><p><strong>∴ Answer: 32/55</strong></p>
Correct Answer: 32/55