Suppose that events M and N are two mutually exclusive events, with P(M) = 0.4 and P (N) = 0.5 . Calculate  P ( N' | M') ?

Given that events M and N are mutually exclusive, this means that they cannot occur at the same time. Therefore, the probability of N given not M (P(N|M')) is simply the probability of N, which is 0.5.

However, it seems like there might be a misunderstanding in the question. If you're asking for the probability of not N given not M (P(N'|M')), this would be a different calculation.

Since M and N are mutually exclusive, P(M and N) = 0. Therefore, P(M' and N') = 1 - P(M or N) = 1 - (P(M) + P(N)) = 1 - (0.4 + 0.5) = 0.1.

The probability of not M, P(M'), is 1 - P(M) = 1 - 0.4 = 0.6.

Therefore, the conditional probability P(N'|M') = P(M' and N') / P(M') = 0.1 / 0.6 = 0.167.