Question:
In a multiple choice questions examination it is known that a student guesses or copies or knows an answer. Probability that he guesses is $1/3$, he copies is $1/6$. If the answer of student is correct, what is the probability that he guessed the answer?
My attempt:
First of all, I calculated the probability that he knows the answer which is equal to $1 -\frac13 - \frac16 = \frac12$.
Now, I think I've to apply Baye's theorem here. But it seems like the information given in the question is not enough to solve the problem.
If I consider $E_1, E_2$ and $E_3$ be the events that he guesses the answer, copies the answer and knows the answer respectively and also let $A$ be the event that his answer is correct, then I've to calculate $P(E_1|A)$.
- $P(E_1) = \frac13$
- $P(E_2) = \frac16$
- $P(E_3) = \frac12$
On applying Baye's theorem, we have that: $$P(E_1|A) = \frac{P(E_1) P(A|E_1)}{P(E_1) P(A|E_1) + P(E_3) P(A|E_3) + P(E_3) P(A|E_3)}.$$
Now I don't have the value of $P(A)$. How would I calculate that? Although $P(A|E_3)$ is $1$ which is easy to find.... if he knows the answer, then surely his answer would be correct but we just can't calculate $P(A|E_1), P(A|E_2) $ and $P(A|E_3)$ without knowing $P(A)$.
Isn't the data provided in the question incomplete?