A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. I have this question as an example in my maths school book. The solution given there is:-
E = the man reports six
P(S1)= Probability that six actually occurs = $\frac{1}{6}$
P(S2)= Probability that six doesn't occur= $\frac{5}{6}$
P(E|S1)= Probability that the man reports six when six has actually occurred = $\frac{3}{4}$
P(E|S2)= Probability that the man reports six when six has not occurred = $1-\frac 3 4=\frac 1 4$
Therefore, by Bayes' Theorem,
$P(S1|E)=\frac{(\frac{1}{6}\cdot\frac{3}{4})}{(\frac{1}{6}\cdot \frac{3}{4})+(\frac{5}{6}\cdot \frac{1}{4})} =\frac{3}{8}  $
I have its solution but my teacher said that the solution given is incorrect and told that the actual solution would be something else:-
$P(S1|E)=\frac{\frac 1 6\cdot\frac3 4}{(\frac 1 6\cdot \frac 3 4)+(\frac 5 6\cdot \frac1 4\cdot\frac1 5)} = \frac 3 4$
So, I want to ask which one is correct.
Thank you.
 A: The difference in solutions comes in the estimation of the probability that the man reports six when six has not occurred.
If the man randomly chooses a number to report when he lies (which seems like a reasonable statement), then the probability he chooses 6 is 1/5. If you multiply your calculation of P(E|S2) by this, you get your teacher's solution.
A: I know I am late, but this might help someone.
The problem arises in the case where the guy didn't get a six, but decides to lie.
So the probability of not getting a six is $\dfrac{5}{6}$.
and the probability that he lies is $\dfrac{1}{4}$.
Fair enough but we still don't know what is the faulty number that he replied with, it might be any number but the actual one.
So the probability that he replies with $6$ is going to be $\dfrac{1}{5}$.
And on solving using Bayes' theorem we get $\dfrac{3}{4}$.
This should be the correct answer for the exact wording of the question as you have provided.
A: Let us define that,
$E_1=A$ speaks truth;      $E_2=A$ tells a lie;       $E= A$ reports a six
Given,
$P (E_1) =\frac{3}{4},~ P (E_2) =\frac{1}{4},~ P (E|E_1) =\frac{1}{6},~  P (E|E_2) =\frac{5}{6}.$
The required probability that actually there were six (by Bayes’) is
$P (E_1|E) = [P (E_1) \cdot P (E|E_1)] / [P (E_1) \cdot P (E|E_1) + P (E_2) \cdot P (E|E_2)] = \frac{3}{8}$.
(Gourav's answer is right)
