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Question:

A boxB1contains1white ball,3red balls and2black balls. Another boxB2contains2white balls,3red balls and4black balls. A third boxB3contains3white balls,4red balls and5black 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 boxB2is5518112618165181116181A boxB1contains1white ball,3red balls and2black balls. Another boxB2contains2white balls,3red balls and4black balls. A third boxB3contains3white balls,4red balls and5black 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 boxB2is5518112618165181116181A boxB1contains1white ball,3red balls and2black balls. Another boxB2contains2white balls,3red balls and4black balls. A third boxB3contains3white balls,4red balls and5black balls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

116181

55181

126181

65181

Solution:

Let A: one ball is white and other is redE1: both balls are from boxB1E2: both balls are from boxB2E3: both balls are from boxB3Required Probability=P(E2A)=P(AE2)⋅P(E2)P(AE1)⋅P(E1)+P(AE2)⋅P(E2)+P(AE3)⋅P(E3)=2C1×3C19C2×131C1×3C16C2×13+2C1×3C19C2×13+3C1×4C112C2×13=1615+16+211=55181