C and D are two events such that P(C) = 0.2, P(D) = 0.6 and P(C|D) = 0.3. Find P(D|C), P(C’ ∩ D’) & P(C’ ∩ D)

Here we use the formula for conditional probability:P(C|D)xP(D)=(𝑃(C∩D))= 0.3x0.6=0.18Note: 𝑃(C∩D)=𝑃(D∩C)=0.18Hence P(D|C)=𝑃(D∩C)/P(C) = 0.18/0.2 = 0.9P(C'∩D') = P(C'|D')xP(D')P(C'|D) = 0.7P(D')=0.4P(C')=0.8P(C∩D')=P(C|D')xP(D') << call this *Also=P(D'∩C)=P(D'|C)xP(C)= *and P(D'|C) = 1-P(D|C)=1-0.9 = 0.1P(D'∩C)=P(D'|C)xP(C)=0.1x0.2 = 0.02 = P(C∩D') 0.02= P(C|D')x0.4P(C|D') = 0.02/0.4 = 0.05P(C'|D') = 1-P(C|D') = 0.95P(C'∩D')=P(C'|D')xP(D') = 0.95 x 0.4 = 0.38P(C’ ∩ D) = P(C'|D) x P(D) = (1-P(C|D)) x P(D) = (1 - 0.3) x 0.6 = 0.6x0.7 = 0.42