I was solving a probability question which is as follow :
A man speaks truth 4 out of 5 times. He throws a die and reports a six. What is the probability that the outcome is actually six ?
The solution involves conditional probability and bayes theorem to solve this. Here it is :
P( actually a six given he reports six) $$= \frac{(\frac{1}{6}.\frac{4}{5})}{ (\frac{1}{6}.\frac{4}{5} + \frac{5}{6}.\frac{1}{5})}$$
However, here's my solution :
The probability that it is actually a six will depend upon whether the man speaks truth or false. Hence , the probability that it is actually six is $\frac{4}{5}$. It shall be independent of whether the experiment involves a die or otherwise , because what he reports and what it is actually , shall solely depends what the man spoke.
Can you give an intuitive explanation that why my solution is wrong ? I don't understand that how a die can change the probability.
P.S.: This question had been already asked with a minor change in data before. However, the answers given were confusing and contradictory. Thanks !
The solution given is wrong. It assumes that if the die comes up $1$ to $5$ and the man lies, he will always report $6$. The statement of the question gives no reason to believe that this is the case.
It might be more reasonable to suppose that when the man lies, he chooses any outcome $1$ to $6$ other than the correct one, with equal probabilities. Then the conditional probability calculation would be $$ \frac{\dfrac{1}{6} \cdot \dfrac{4}{5}}{\dfrac{1}{6} \cdot \dfrac{4}{5} + \dfrac{5}{6} \cdot \dfrac{1}{5} \cdot \dfrac{1}{5}} = \dfrac{4}{5} $$ in agreement with your solution, that this is just the probability that the man is lying.
However, as far as we know, the man could lie by saying $2023$ or $\pi$ or anything else. The question is just not well stated.