In a test, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices, only one answer being correct. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct, given that he copies it, is $\frac{1}{8}$
What is the probability that he knew the answer to the question, given that he correctly answers it?
From the text I identified $3$ events regard to the same experience.So the sum of the $3$ probabilities must be $1$.It's known that:
$P(A)=\frac{1}{3}$
$P(B)=\frac{1}{6}$
So,
$P(C)+\frac{1}{3}+\frac{1}{6}=1$
$P(C)=\frac{1}{2}$, this is the probability of knowing the answer.
It's also known that $P(D|B)=\frac{1}{8}$. $P(D)$ is the probability of the question is correctly answered.
The problem ask about $P(C|D)$. From the knowledge that $P(D|B)=\frac{1}{8}$, $P(B)=\frac{1}{6}$ and by the definition of conditional probability, it's known that
$P(D \cap B)=\frac{1}{8} \cdot \frac{1}{6}$. This proves that $B$ and $D$ are indepentend events. And if the $P(B)$ it's known, the $P(D)$ must be $\frac{1}{8}$.
Now, using the conditional probability definition, one can find $P(C|D)$.But if $D$ and $B$ were independent, and $C$ and $B$ are events of the same experience, than $C$ and $D$ must also be independents. So $P(C|D)=P(C)=\frac{1}{2}$
Is my thought right?