On a question chosen at random, what is the probability that the student answers it correctly? 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.
$\begin{align}
P(\text{know correct answer}|\text{answer correctly}) & = P(\text{know correct answer}|\text{answer correctly}) \\
& =\frac{P(\text{know correct answer and answer correctly})}{P(\text{answer correctly)}}. \\
\end{align}$
 A: 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.
A: Consider a typical run of 100 questions.
How many will he know the answer for?  Obviously, he gets all of these right.
How many are left to guess about?  How many of these will he get right?
How many correct all together from the entire 100 question run?
Your logic for the second question is correct, given the correct numbers to insert...
A: Please go through these to understand the Concept clearly.
Question 1:
In answering a question on a multiple-choice test, an examinee either
knows the answer (with probability p), or he Guesses (with probability 1 - p).
Assume that the probability of answering a question correctly is unity for an examinee who knows the answer and 1/m for the examinee who guesses, where m is the number of multiple-choice alternatives. Supposing an examinee answers a question correctly, what is the probability that he really knows the answer?
Solution :
MCQ : m options.
P(KNOWS the correct answer) : p
P(GUESSES the correct answer) : (1 - p)
The probability of answering a question correctly is unity for an examinee who knows the answer.
A = The examinee answers CORRECTLY.
Let K = The examinee KNOWS the answer.
Then , $P(\frac{A}{K}) = 1$
The probability of answering a question correctly is 1/m for the examinee who GUESSES, where m is the number of multiple-choice alternatives.
A = The examinee answers correctly.
Let G = The examinee GUESSES the answer.
Then, $P(\frac{A}{G}) = \frac{1}{m}$
Then, the conditional probability that a man knew the answer to a question, given that he has Correctly answered it, is equal to $P (K | A  ) = P( \frac{\text{Man knew the answer to the Question}}{\text{He has correctly answered it}}) = P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question OR He Guessed the answer }} )= P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question + He Guessed the answer }} ) =\frac{p(1)}{p(1) + (1-p)\frac{1}{m}} = \frac{mp}{mp + 1- p}$
Now If we add 1 more condition of Copying. Then, Let us look at this Question
Question 2:
In a test, an examinee, either Guesses Or Copies Or Knows the answer for multiple-choice test having 4 options of which only 1 is correct.The probability that he makes a guess is 1/3 and the probability for copying is 1/6. The probability that his answer is correct given that he copied it is 1/8. Prove that The probability that he knew the answer, given that his answer is correct is 24/29.
Solution :
Let, C be the probability that he will COPY the answer.
C = $\frac{1}{6}$
A = The examinee answers CORRECTLY.
Then, $P(Correct|Copy)  = P(A|C) =(\frac{1}{8})$
The probability of answering a question correctly is 1/m for the examinee who GUESSES, where m is the number of multiple-choice alternatives.
A = The examinee answers correctly.
Let G = The examinee GUESSES the answer. = 1/3 
Then, $P(\frac{A}{G}) = \frac{1}{m} = \frac{1}{4}$
Let K = The examinee KNOWS the answer.
Then  $K = 1 - (G+C) = 1 - (\frac{1}{6} + \frac{1}{3}) = \frac{1}{2}$
Here also, we will say: the Probability that his answer is correct given that he KNOWS the answer => $P(A|K) = 1 $. 
The probability that he knew the answer, given that his answer is correct  =
$ P( \frac{\text{Man knew the answer to the Question}}{\text{He has correctly answered it}}) = P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question OR He Guessed the answer OR He Copied the correct answer}} )= P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question + He Guessed the answer + He Copied the correct answer}} )  => P(K|A) =  \frac{P(K).P(A|K)}{P(K).P(A|K) + P(G).P(A|G) + P(C).P(A|C)} =  \frac{P(K).(1)}{P(K).(1) + P(G).(\frac{1}{options}) + P(C).(\frac{1}{8})} =   \frac{\frac{1}{2}.(1)}{\frac{1}{2}.(1) + \frac{1}{3}.(\frac{1}{4}) + \frac{1}{6}.(\frac{1}{8})} =  \frac{24}{29}$
