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 3 4)+(\frac 5 6\cdot 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.
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.
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)