An urn contains 10 white and 3 black balls. Another urn contains 3 white and 5 black balls. Two are drawn from first and put into the second urn and then a ball is drawn from the latter. Find the probability that it is a white ball.

Options

A. `pgtq`
B. `pltq`
C. `41p=22q`
D. `22p=41q`

Answer

B, C

Explanation

`E_(1)to` Two white balls are tranferred
`E_(2)to` Two black balls are transferred
`E_(3)to` One white & one black ball is transferred
`p=P(E_(1)).P(W|E_(1))+P(E_(2)).P(W|E_(2))+P(E_(2)).P(W|E_(3))`
`(( .^(4)C_(2))/( .^(7)C_(2)))((4)/(9))+(( .^(3)C_(2))/( .^(7)C_(2)))((2)/(9))+(( .^(4)C_(1) .^(3)C_(1))/( .^(7)C_(2)))((3)/(9))`
`=((6)/(21))((4)/(9))+((3)/(21))((2)/(9))+((12)/(21))((3)/(9))`
`=(24+6+36)/(186)=(66)/(189)`
`q=P(E_(1)).P_(B|E_(1))+P_(E_(2)).P_(B|E_(2))+P(E_(3))P_(B|E_(3))`
`=(( .^(4)C_(2))/( .^(7)C_(2)))((5)/(9))+(( .^(3)C_(2))/( .^(7)C_(2)))((7)/(9))+(( .^(3)C_(1) .^(4)C_(1))/( .^(7)C_(2)))((6)/(9))`
`=((6)/(21))((5)/(9))+((3)/(21))((7)/(9))+((12)/(21))((6)/(9))`
`=(30+21+72)/(189)=(123)/(189)`