Correct Answer - A::D
`P(A^(c)) = 0.3 " " ["given"]`
`rArr P(A) = 0.7`
`P(B) = 0.4 " "["given"]`
`rArr P(B^(c)) = 0.6 " and "P (A nn B^(c)) = 0.5 " " ["given"]`
Now, `P(A uu B^(c)) =P(A) +P(B^(c)) - P(A nn )^(c))`
` = 0.7 +0.6 -0.5 = 0.8`
`therefore P[B//(A uu B^(c)] = (P{B nn (A uu B^(c))})/(P(A uu B^(c)))`
`=(P{(B nn A) uu (B nn B^(c))}}/(0.8) = (P{(B nn A) uu phi})/(0.8) = (P(B nn A))/(0.8)`
`=(1)/(0.8) [P(A) - P(A nn B^(c))]`
` = (0.7 -0.5)/(0.8) = (0.2)/(0.8) = (1)/(4)`