A student takes a multiple choice exam in which each question has $5$ possible answers. The student knows the answer to $70\%$ of all questions. If he doesn't know the answer to any question, he guesses it randomly.
(i) What is the probability that the student chooses the correct answer for a given question?
(ii) If the student chooses the correct answer for a given question, what is the probability that he knew the answer?
In my solution I'm unsure about part (ii). My solution:
Let us define some events. $$S := \text{The student knows the answer}. $$
$$ S^{c} := \text{The student does not know the answer}. $$
$$A := \text{The students chooses the right answer.} $$
Then,
$$ P(S) = 70\% $$
$$ P(S^{c}) = 30\% $$
Part (i) asks us to find $P(A)$. That is the probability of knowing the answer $+$ the probability of guessing the correct answer.
$$ P(A) = 70\% + \Bigl( \frac{1}{5} \Bigr) 30\% = 76\% $$
Part (ii) asks us to find $P(A|S)$. By the definition of conditional probability,
$$ P(A|S) = \frac{P(A \cap S)}{P(S)} $$
$ P(A \cap S) \neq P(A)P(S) $ as $A$ and $S$ are not independent. So the "difficult" part is finding $ P(A \cap S) $. My attempt:
$$ P(A \cap S) = 1 – P(A \cap S)^c $$
$$ (A \cap S)^c = \{x : x \notin A ~and~ x \notin S\} = A^c \cup S^c $$
$$ P(A^c \cup S^c) = P(A^c) + P(S^c) = 24\% + 30\% = 54\% $$
Therefore
$$ P(A \cap S) = 1 – P(A \cap S)^c = 1 – P(A^c \cup S^c) = 1 – 0.54 = .46 = 46\% $$
Finally
$$ P(A|S) = \frac{46\%}{70\%} = \frac{46}{70} = \frac{23}{35} $$
Best Answer
Bayes theorem gives the answer: $$ P(S|A) = \frac{P(A|S)P(S)}{P(A|S)P(S)+P(A|S^c)P(S^c)}=\frac{1*0.7}{1*0.7 + 0.2*0.3}=.921 $$