The events Y and Z are mutually exclusive and exhaustive: P(Y) = 0.4 and P(Z) = 0.6. If the probability of X given Y is 0.9, and the probability of X given Z is 0.1, what is the unconditional probability of X? A) 0.40. B) 0.42. C) 0.33. D) 0.66. Answer is C. Can anyone explain why and how they got the answer? Thanks
0.9 = P(X | Y) = P(X & Y)/P(Y) => P(X&Y) =0.36 0.1 = P(X|Z) = P(X&Z)/P(Z) => P(X&Z) = 0.06 P(X) = P(X&Y) + P(X&Z) = 0.42 = answer B) not answer C).
I got B as well.
Prob(X/Y) = Prob(X and Y) / Prob (Y) so, Prob(X and Y) = Prob(X/Y) * Prob(Y) Prob(X and Y) = 0.9*0.4 = 0.36 Similarly, Prob(X/Z) = Prob(X and Z) / Prob (Z) so, Prob(X and Z) = Prob(X/Y) * Prob(Y) Prob(X and Z) = 0.1*0.6 = 0.06 Since, It’s given that the events Y and Z are mutually exclusive and exhaustive events. Prob of both occuring simultaneously is 0 i.e. Prob (Y and Z) = 0 unconditional P(X) = condition probablity of X in all possible states of the world so P(X) = P(X/Y) + P(X/Z) - P(Y and Z) P(X) = 0.36 + 0.06 = 0.42 So the ans should be ‘B’ hope I’m correct at my probability concepts… - Dinesh S
I understand up to 0.9 = P(X | Y) = P(X & Y)/P(Y) => P(X&Y) =0.36 0.1 = P(X|Z) = P(X&Z)/P(Z) => P(X&Z) = 0.06 and this is because of the general rule of mutliplication… but how do you know that P(X) = P(X&Y) + P(X&Z) Am I missing a another rule here?
Thanks so much Dinesh. It makes sense now. I was missing that last part.
The correct answer is B by the way. But I responded C which was incorrect.
Philly1616, the easiest way to solve the problem is to use the law of total probability directly: P(X) = P(X|Y)*P(Y) + P(X|Z)*P(Z) = .9*.4+.1*.6 = .42