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.  
Thanks in advance.
 A: 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 $8/9$.
A: 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.
A: Please go through these to understand the concept :
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}$
