The die.. problem

We just need to justify the person is telling truth or not, and $Pr($ this person tells truth $) = \boxed{\cfrac{3}{4}}$

Some may use below table to find out the solution:

They suggest the prob. $= \cfrac{\cfrac{3}{24}}{\cfrac{3}{24}+\cfrac{5}{24}}=\cfrac{3}{8}$

But for the case the person tell lies when the die is NOT rolled 6, there is only $\cfrac{1}{5}$ chance that the person reports the rolled number is $6$

So the prob. should be $= \cfrac{\cfrac{3}{24}}{\cfrac{3}{24}+\cfrac{5}{24}\cfrac{1}{5}}=\boxed{\cfrac{3}{4}}$

I think if the probability the person tells the truth/lies is independent of the roll; Then, if said person says "its a 6", the probability of the roll being a 6 is equivalent to the probability that they tell the truth for the outcome of a given roll ( i.e it doesn't depend on how probable the roll "being a 6" is ).

Thus; $P = \frac{3}{4}$