The Liar (Baye's theorem question)

first find probability of reporting six as truth-

P1= 1/6x3/4=3/24 1/6 - probability of appearance of six 3/4 - probability of speaking truth thus their product gives probability of their simultaneous occurrence

now find probability of reporting six as false-

for this, first either of 1,2,3,4,5 must have appeared on the die, then we multiply by 1/4 as the person must be speaking false, and then as false, he can speak any 5 numbers except the one that appeared on the die and of these 5, the probability of him speaking 6 will be 1/5.

Thus P2= 5/6x1/4x1/5=1/24

Thus, he speaks truth 6 with probability 3/24 and false 6 with probability 1/24. So probability of truth 6 when it is given that he spoke 6 is (3/24)/(3/24+1/24)= 3/4

By Bayes's Theorem: $\begin{aligned} P(\text{die shows 6}\mid\text{man says 6}) &= \frac{P(\text{man says 6}\mid\text{die show 6})\times P(\text{die shows 6})}{P(\text{man says 6})} \\ &= \frac{\frac{3}{4}\times\frac{1}{6}}{\frac{1}{6}\times\frac{3}{4} + \frac{5}{6}\times\frac{1}{4}\times{\color{#3D99F6}\frac{1}{5}}} && \color{#3D99F6}\text{Assuming he chooses randomly when he lies} \\ &= \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{24}} \\ &= \frac{24}{8(3+1)} \\ &= \boxed{\frac{3}{4}} \end{aligned}$

If your text said something different, it's possible they were assuming the man will always say "six" when he lies (if the roll wasn't actually a six), in which case the answer would be $\frac{3}{8}$ .

Or there was a mistake in the solution (it happens).