Question:
There are two bags. The first bag contains 3 white and 4 red balls and the second bag contains 4 white and 3 red balls. Two balls are drawn at random from the first bag and are put into the second bag without noticing their colours. Then two balls are drawn from the second bag. If the probability that one ball is white and the other is red is \( \frac{x}{y} \), where \( \frac{x}{y} \) is in the simplest form, then value of \( (2x-y) \) is ________
There are two bags. The first bag contains 3 white and 4 red balls and the second bag contains 4 white and 3 red balls. Two balls are drawn at random from the first bag and are put into the second bag without noticing their colours. Then two balls are drawn from the second bag. If the probability that one ball is white and the other is red is \( \frac{x}{y} \), where \( \frac{x}{y} \) is in the simplest form, then value of \( (2x-y) \) is ________
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Condition on the colours of the two balls moved from bag 1 to bag 2 (WW, WR, RR) and use the total probability theorem.
Updated On: Nov 12, 2026
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Correct Answer: 4
Solution and Explanation
Two balls from bag 1 (7 balls): P(WW) = 3/21, P(WR) = 12/21, P(RR) = 6/21.
Bag 2 then has 9 balls; the chance of drawing one white and one red is (W x R) / C(9,2), with C(9,2) = 36.
If WW moved: 6 W, 3 R, probability = 18/36 = 1/2.
If WR moved: 5 W, 4 R, probability = 20/36 = 5/9.
If RR moved: 4 W, 5 R, probability = 20/36 = 5/9.
Total = (3/21)(1/2) + (12/21)(5/9) + (6/21)(5/9) = (1/21)(3/2 + 10) = 23/42.
So \( x=23, y=42 \) (co-prime) and \( 2x-y = 46-42 = 4 \).
Answer: 4
Bag 2 then has 9 balls; the chance of drawing one white and one red is (W x R) / C(9,2), with C(9,2) = 36.
If WW moved: 6 W, 3 R, probability = 18/36 = 1/2.
If WR moved: 5 W, 4 R, probability = 20/36 = 5/9.
If RR moved: 4 W, 5 R, probability = 20/36 = 5/9.
Total = (3/21)(1/2) + (12/21)(5/9) + (6/21)(5/9) = (1/21)(3/2 + 10) = 23/42.
So \( x=23, y=42 \) (co-prime) and \( 2x-y = 46-42 = 4 \).
Answer: 4
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