<p>A bag has 10 balls. Six balls are drawn in an attempt and replaced. Then another draw of 5 balls is made from the bag. The probability that exactly two balls are common to both the draw is</p>
Step-by-Step Solution
Key Concept: This is sampling with replacement problem. We need to find the probability that exactly 2 balls from the second draw of 5 balls are among the 6 balls drawn in the first attempt.
<p><strong>Step 1:</strong> Recognize the setup: First draw takes 6 balls (they are replaced), second draw takes 5 balls. We want exactly 2 balls common to both draws.</p><p><strong>Step 2:</strong> In the second draw of 5 balls, we need exactly 2 to come from the first 6 balls and exactly 3 to come from the remaining 4 balls.</p><p><strong>Step 3:</strong> Number of ways to choose 2 balls from the 6 that were drawn first: C(6,2)</p><p><strong>Step 4:</strong> Number of ways to choose 3 balls from the 4 that were NOT drawn first: C(4,3)</p><p><strong>Step 5:</strong> Total ways to draw 5 balls from 10: C(10,5)</p><p><strong>Step 6:</strong> Probability = [C(6,2) × C(4,3)] / C(10,5)</p><p><strong>Step 7:</strong> Calculate: [15 × 4] / 252 = 60/252 = 5/21</p><p>∴ Answer: A</p>
Correct Answer: A