Bag A contains x red balls and 6 blue balls. Bag B

Solution

ATQ, Total balls in Bag A = x + 6. Total balls in Bag B initially = 9. Case 1: A red ball is transferred from A. Probability of picking red from A = x/(x + 6). Then Bag B will have 6 red and 4 blue (10 balls total). Probability of then drawing red from B = 6/10 = 3/5. Case 2: A blue ball is transferred from A. Probability of picking blue from A = 6/(x + 6). Then Bag B will have 5 red and 5 blue (10 balls total). Probability of then drawing red from B = 5/10 = 1/2. Total probability (P) of drawing red from B: P = [x/(x + 6)] × (3/5) + [6/(x + 6)] × (1/2). Given P = 27/50: [x/(x + 6)] × 3/5 + [6/(x + 6)] × 1/2 = 27/50 Take common denominator (x + 6): (1/(x + 6)) [ (3x)/5 + 3 ] = 27/50 ⇒ (3x/5 + 3) / (x + 6) = 27/50 Multiply both sides by (x + 6): 3x/5 + 3 = 27/50 (x + 6) Multiply both sides by 50: 50(3x/5 + 3) = 27(x + 6) 10(3x + 5 × 3)? Careful: 3x/5 + 3 = (3x + 15)/5 50 × (3x + 15)/5 = 10(3x + 15) = 30x + 150. Right side: 27(x + 6) = 27x + 162. So: 30x + 150 = 27x + 162 3x = 12 x = 4. Answer: x = 4 (Bag A has 4 red and 6 blue balls).