3 candidates for president, labeled A, B,C lie 5%, 10%, 15% of the time respectively, tell half truths 40%, 45%, 50% of the time respectively and tell the truth 55%, 45%, and 35% respectively. Assume A tells twice as many statements as B and 3 times as many as C.

If a lie was told by a candidate, what is the probability it was from A? What percentage of all statements are half-truths?

OK..

So I proceeded with using the

conditional probability formula: P(A|B) = P(AB)/P(B) but expanded it using the idea of Bayes Theorem:

P(A|E) = P(A)P(E|A)/P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)

where A is the one we want and E is some subset.

So I got (1/3)(5/100)/(1/3)(5/100)+(1/3)(10/100)+(1/3)(15/100)

This will equal .1666666 but I times it by two since A issues twice as many statements as B. So the answer is 1/3. But I'm not sure if this is correct interpretation.

The second part of the question simply confuses me, I don't know where to proceed with that one.