Q) A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually six.

A) Since he has already reported that it's a 6, either he is lying about it or it is actually a six.
That is, either it is truly a six & he's telling the truth, OR it's a non-six & he's lying by saying that it's a six.

Net Probability = (prob. of actual 6 & truth) / [prob. of actual 6 & truth + prob. of non 6 & lies ]
= (3/4)(1/6) / [3/4*1/6 + 1/4*5/6]
= 3/8

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