A man is known to speak the 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 tried in this manner that as speaking truth as the event then $P(T) = \frac{3}{4}$ and dice to fall as 6 the event denoted as D so $P(D=6) = \frac{1}{6}$ and dice and speaking truth is an independent event so we can write as $$P(D|T) = P(D)P(T) = \frac{1}{6} \times \frac{3}{4} = 1/8$$ Is this the correct answer?
On
Your answer is not correct because of your small misundertanding of independence, since if ypu say that events $T$ and $D$ are independents, then that means: $$P(D|T)=P(D)$$ But at this time $D$ and $T$ are not independents because you dont have the outcome landed by the die but a stament wich is true with probability 3/4. So you need to use Bayes's Formula here as highlighted by @tommik.
No, they are not independent. He can claim Six when he get actually six or claim six when he get another number. Using Bayes' theorem:
$$P(\text{Six}|\text{Claimed Six})=\frac{\frac{1}{6}\frac{3}{4}}{\frac{1}{6}\frac{3}{4}+\frac{5}{6}\frac{1}{4}}=\frac{3}{3+5}=\frac{3}{8}$$