Given in the question: A is known to speak truth 3 times out of 7 times. A throws a dice and reports that it is 1. calculate: The probability that it is actually 1. Answer: Probability that it was actually a 1 = \frac{1}{6} Probability that he said the truth = \frac{3}{7} Combined probability that he said the truth when it was actually a 1 = \frac{1}{6} *\frac{ 3} {7} =\frac{1}{14}. The another possible scenario: Probability that it wasn't a 1 = \frac{5}{6} Probability that he lied = \frac{1}{7} Important Note: If the man randomly chooses a number to report when he lies, then the probability he chooses 1 is \frac{1}{5}. Suppose the outcome was 4. Given that he is going to lie, he could've used any of the remaining 5 numbers. So, the probability that he chooses 1 in order to lie is \frac{1}{5}. Combined probability of this scenario = \frac{ 5}{6} *\frac{1}{7} *\frac{1}{5} = \frac{1}{42} Hence required probability =\frac{\frac{1}{14}} {\frac{1}{14} +\frac{1}{42}} = 3/4 or 75%.