Answer: 51%

The correct answer is C (51%). **Explanation:** We need to calculate P(2nd ball is red) using conditional probability and the law of total probability. **Step 1: Calculate P(1st ball is red)** - Probability of selecting Bag 1 = 1/2 - Probability of drawing red from Bag 1 = 5/10 = 1/2 - Probability of selecting Bag 2 = 1/2 - Probability of drawing red from Bag 2 = 3/5 P(1st red) = (1/2 × 1/2) + (1/2 × 3/5) = 1/4 + 3/10 = 0.25 + 0.30 = 0.55 **Step 2: Calculate P(1st ball is white)** P(1st white) = 1 - P(1st red) = 1 - 0.55 = 0.45 **Step 3: Calculate P(2nd red | 1st red)** If the first ball is red, we draw the second ball from the same bag. Case 1: First ball red from Bag 1 - Probability: (1/2 × 1/2) / 0.55 = 0.25/0.55 = 5/11 - After drawing one red from Bag 1: 4 red, 5 white remain - P(2nd red | Bag 1, 1st red) = 4/9 Case 2: First ball red from Bag 2 - Probability: (1/2 × 3/5) / 0.55 = 0.30/0.55 = 6/11 - After drawing one red from Bag 2: 2 red, 2 white remain - P(2nd red | Bag 2, 1st red) = 2/4 = 1/2 P(2nd red | 1st red) = (5/11 × 4/9) + (6/11 × 1/2) = (20/99) + (6/22) ≈ 0.2020 + 0.2727 = 0.4747 **Step 4: Calculate P(2nd red | 1st white)** If the first ball is white, we draw the second ball from the other bag. Case 1: First ball white from Bag 1 - Probability: (1/2 × 1/2) / 0.45 = 0.25/0.45 = 5/9 - Then draw from Bag 2 (original composition: 3 red, 2 white) - P(2nd red | Bag 2) = 3/5 = 0.6 Case 2: First ball white from Bag 2 - Probability: (1/2 × 2/5) / 0.45 = 0.20/0.45 = 4/9 - Then draw from Bag 1 (original composition: 5 red, 5 white) - P(2nd red | Bag 1) = 5/10 = 0.5 P(2nd red | 1st white) = (5/9 × 3/5) + (4/9 × 1/2) = (15/45) + (4/18) = 1/3 + 2/9 = 3/9 + 2/9 = 5/9 ≈ 0.5556 **Step 5: Calculate total probability using law of total probability** P(2nd red) = P(2nd red | 1st red) × P(1st red) + P(2nd red | 1st white) × P(1st white) = 0.4747 × 0.55 + 0.5556 × 0.45 = 0.2611 + 0.2500 = 0.5111 ≈ 51.11% This matches option C (51%).

Author: Nikitesh Somanthe