A gambler throws a die and tells the score. He tells truth 3/4 of the times that he speaks and randomly lies 1/4 of the times that he speaks. If he says it is a six, what is the probability that he actually got a six?

OK, I'll jump in with conditional probability.

probability of having a 6 and telling the truth about it =

(probability of having a 6)

*(probability of telling truth)

= (1/6)*(3/4) = 3/24

Probability of not having a 6 but claiming that you do =

(probability of not having a 6)

*(probability of lieing)

*(probability that the number claimed, which must differ from actual roll, is 6) =

(5/6)*(1/4)*(1/5) = 1/24

Answer to problem =

(3/24)/(3/24 + 1/24) = 3/4