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Question

A box B1 contains 1 white ball, 3 red balls, and 2 black balls. Another box B2 contains 2 white balls, 3 red balls, and 4 black balls. A third box B3 contains 3 white balls, 4 red balls, and 5 black balls.
If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is

A
116182
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B
126181
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C
65181
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D
55181
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Solution

The correct option is D 55181
Let
A: one ball is white and other is red
E1: both balls are from box B1
E2: both balls are from box B2
E3: both balls are from box B3

Here, P(required)=P(E2A)
=P(AE2).P(E2)P(AE1).P(E1)+P(AE2).P(E2)+P(AE3).P(E3)
= 2C1× 3C1 9C2×13 1C1× 3C1 6C2×13+ 2C1× 3C1 9C2×13+ 3C1× 4C1 12C2×13
=55181

Alternate Solution:

Probability of drawing a white and a red ball from bag B1 is P(B1)=1C1×3C16C2=15
Probability of drawing a white and a red ball from bag B2 is P(B2)=2C1×3C19C2=16
Probability of drawing a white and a red ball from bag B3 is P(B3)=3C1×4C112C2=211
Therefore the probability that the red and the white ball drawn are from the bag B2 is P(B2)P(B1)+P(B2)+P(B3)=55181

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