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
Probability doesn't lie either (spoiler)
