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%.