# Conditional Probability Problem About a Quiz in Which Every Question Has 4 Answers

Imagine that a quiz has some questions and every question has 1 correct answer between 4 choices. The probability that the student knows the answer of each question is $\frac{2}{3}$.

What's the probability that if the student answered a question correctly, he knew the answer?

Note: I even know that the answer of this question is $\frac{8}{9}$ but i don't know why! My problem is that i don't know which part should be the condition and also i'm not sure that the question wants the probability of an intersection or a conditional probability.

I have three events, $C$ is the event that the problem was correct, $K$ is the event that the student knew the answer, and $G$ is the event that the student guessed. Now, we proceed mechanically, and the problem becomes \begin{align*} P(K|C)&= \frac{P(K,C)}{P(C)}\\ &=\frac{P(C|K)P(K)}{P(CG)+P(C\bar G)}\\ &=\frac{P(C|K)P(K)}{P(C|G)P(G)+P(C|\bar G)P(\bar G)}\\ &=\frac{(1)(2/3)}{(1/4)(1/3)+(1)(2/3)}\\ &=\frac{8}{9} \end{align*} where $\bar G$ means that the student did not guess, which means the student knew the answer. Further, if the student guesses, then there is a $1/4$ chance that the answer is correct. If the student does not guess or knows the answer, then there is a $100\%$ chance that the answer is right.
Recall from conditional probability that $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Here, your events are knowing the answer and answering correctly respectively. You are already given the probability of knowing the answer and answering correctly, precisely because the event of knowing the answer is a subset of answering correctly (if you know the answer, you will answer correctly), so has the probability of just knowing the answer. So you just need to find the probability of answering correctly. This can happen in two mutually exclusive ways: either the student knows the answer with probability $2/3$, in which case he/she answers correctly guaranteed, or does not know it with probability $1/3$, in which case must guess and has a probability of $1/4$. Using that information, find the probability, do the division and you'll get $8/9$.