`Aa n dB` are two candidates seeking admission in ITT. The probability that `A` is selected is 0.5 and the probability that `Aa n dB` are selected is at most 0.3. Is it possible that the probability of `B` getting selected is 0.9?

Explanation

P(A) = 0.5, `P(A nn B) le 0.3`
So, `P(A uu B) = P(A) + P(B) - P(A nn B)` gives
`P(B) = P(A uu B) - P(A) + P(A nn B) le 1 + 0.3 - 0.5 = 0.8 " "[ because P(A uu B) le 1]`
Hence, P(B) = 0.9 is not possible