A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls and 4 black balls. A third box `B_(3)` contains 3 white balls, 4 red balls, and 5 black balls.
If 2 balls are drawn (without replecement) 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 `B_(2)` is

Options

A. `116//182`
B. `126//181`
C. `65//181`
D. `55//181`

Answer

D

Explanation

Let A: one ball is white and other is red
`E_(1):` both balls are from box `B_(1),`
`E_(2): ` both balls are from box `B_(2),`
`E_(3):` both balls are form box `B_(3)`
Hence, P (required) `=P((E_(2))/(A))`
`=(P((A)/(E_(2))).P(E_(2)))/(P((A)/(E_(1))).P(E_(1))+P((A)/(E_(2))).P(E_(2))+P((A)/(E_(3))).P(E_(3)))`
`=((""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3)/((""^(1)C_(1)xx""^(3)C_(1))/(""^(6)C_(2))xx1/3+(""^(2)C_(1)xx""^(3)C_(1))/(""^(9)C_(2))xx1/3+(""^(3)C_(1)xx""^(4)C_(1))/(""^(12)C_(2))xx1/3)`
`(1/6)/(1/5+1/6+2/11)=55/181`