I'm really confused about this question. I appreciate your help.
A student takes a multiple choice exam where each question has five possible answers. He answers correctly if he knows the answer, otherwise he guesses at random. Suppose he knows the answer to $70$% of the questions.
Question 1
On a question chosen at random, what is the probability the student answers it correctly?
We know $P(\text{know the answer}) = 0.7$. On a question chosen at random, he either knows the answer or he doesn't. If he does, then $P(\text{answer correctly}|\text{know correct answer})= 1$. If he doesn't, then $P(\text{answer correctly}|\text{doesn't know correct answer})= 0.7(0.3)^4$. I don't know how to continue the answer.
Question 2
Given that he did answer correctly, what is the probability that he actually knew the correct answer?
All I can think of is the following conditional probability.
$\begin{align}
P(\text{know correct answer}|\text{answer correctly}) & = P(\text{know correct answer}|\text{answer correctly}) \\
& =\frac{P(\text{know correct answer and answer correctly})}{P(\text{answer correctly)}}. \\
\end{align}$
Best Answer
For the first one, drawing a probability tree might help.
At first, you have the two branches of whether he knows the answer (0.7) and he doesn't know the answer (0.3).
If he knows the answer, then he answers correctly (1), so that, the probability that he knows and answer AND answers correctly is (0.7*1 =) 0.7.
If he doesn't know the answer, he has a probability of (1/5 =) 0.2 of getting the right answer, so that the probability that he doesn't know the answer AND answers correctly becomes: (0.3*0.2 =) 0.06
The rest should be a little easier.