Person speaking truth and a dice is thrown , probability problem.

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A person speaks the truth 8 times out of 10 times. A die is rolled. He reports that is was 5. What is the probability that it was actually 5.

A = A speaks the truth. F = 5 appears

P(A) = 8/10 ,P(F) = 1/6

$$P(F|A) = \frac{P(A|F).P(F)}{P(A|F).P(F) + P(-A|F).P(F)}$$

Is this the correct way of doing this? I am really confused. Correct me if I am wrong.

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I'll assume that if the person lies, he randomly selects among the $5$ wrong numbers with equal probability. In that case, the probability that the roll really was a $5$ is $0.8$.

Let's perform $30$ trials. In $5$ of those trials, the die will come up $5$ and in $4$ of those $5$ trials, the person will truthfully report $5$.

In the other $25$ trials, the die will come up some number other than $5$. In $5$ of those $25$ trials, the person will lie. In $1$ of those $5$ trials, the lie will be a false report that the die roll was a $5$.

Thus, in $5$ trials, the person will report that the die roll was a $5$. He'll be telling the truth in $4$ of those $5$ trials. Thus, given that the report is a $5$, the probability that the die roll was actually a $5$ is $0.8$.