A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually six is 1. 1/8 2. 2/8 3. 3/8 4. 1/2

This is a case of multiple probability Probability of him saying the truth is : 3/4 Probability of him gettin a 6 on roll : 1/6 Multiplying 3/4 * 1/6 = 1/8 Hence the correct answer is 1/8

IMO 3/8 The thing is if it is not six he can lie that it is six or if it is six then he can tell truth if it is not six and he told lie => 5/6*1/4 = A(say) if it is six and he told truth => 1/6*3/4 = B(say) probability that it is six and he told truth is (1/6*3/4)/((1/6*3/4)+(5/6*1/4)) just like A/(A+B) => 3/8 user123321

user123321, if it is not six and he told lie => 5/6*1/4 = A(say) The question asks for the probability that it is a six- "He throws a die and reports that it is a six. The probability that it is actually six is " . So Case A is not valid isn't it, why do we need to worry if he lied.

Step1: Probability of getting a six in the throw of a die=1/6 Step2: Truth = 3/4 So the combined probability = 3/4*1/6=1/8

I agree with user123321, we need to consider both cases as the man has already reported the number as 6. But we are interested only in knowing if he said the truth. P(getting 6/ reporting 6) = P(getting 6 and reporting six)/ [P(getting 6 and reporting six) + P(getting any other number expect 6 and reporting 6)] = (1/6 * 3/4) /[(1/6 * 3/4) + (5/6 * 1/4)] = 3/8