Conditional probability and independent events. 
In a test, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices, only one answer being correct. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct, given that he copies it, is $\frac{1}{8}$

What is the probability that he knew the answer to the question, given that he correctly answers it?
From the text I identified $3$ events regard to the same experience.So the sum of the $3$ probabilities must be $1$.It's known that:
$P(A)=\frac{1}{3}$
$P(B)=\frac{1}{6}$
So,
$P(C)+\frac{1}{3}+\frac{1}{6}=1$
$P(C)=\frac{1}{2}$, this is the probability of knowing the answer.
It's also known that $P(D|B)=\frac{1}{8}$. $P(D)$ is the probability of the question is correctly answered.
The problem ask about $P(C|D)$. From the knowledge that $P(D|B)=\frac{1}{8}$, $P(B)=\frac{1}{6}$ and by the definition of conditional probability, it's known that
$P(D \cap B)=\frac{1}{8} \cdot \frac{1}{6}$. This proves that $B$ and $D$ are indepentend events. And if the $P(B)$ it's known, the $P(D)$ must be $\frac{1}{8}$.
Now, using the conditional probability definition, one can find $P(C|D)$.But if $D$ and $B$ were independent, and $C$ and $B$ are events of the same experience, than $C$ and $D$ must also be independents. So $P(C|D)=P(C)=\frac{1}{2}$
Is my thought right?
 A: First, some warm up:
Let's define our events at the start:
$\ \ \ D$ is the event that the student answers correctly.
$\ \ \ A$ is the event that the student guesses the answer.
$\ \ \ B$ is the event that the student copies the answer.
$\ \ \ C$ is the event that the student knows the answer.
Let's also write down what we know:
$$\textstyle
P(A)={1\over3},\quad P(B)={1\over 6},\quad P(D\mid B)={1\over 8} 
.
$$
Also note
$$\textstyle
 P(D\mid C)=1,
\quad P(D\mid A)={1\over 4},\quad  P( C) =1-{1\over3}-{1\over6}={1\over2}.
$$

Now on to the problem proper:
You want to find $P(C\mid D)$.
$C$ and $D$ are not independent. We  have to use the basic formula defining conditional probabilities:
$$\tag{1}
P(C\mid D) ={P(C\cap D)\over P(D)}.
$$
To find $P(C\cap D)$, we use the basic formula again (though it's usually called the multiplication principle when used this way): 
$$
P(C\cap D) =P(C)P(D\mid C).
$$
We know $P(C)={1\over2}$ (as you calculated); and, if we're given that the student knows the answer, it follows that in this case that the probability that the student answers correctly is 1. Thus 
$$\textstyle\tag{2}P(C\cap D) = {1\over2}\cdot 1={1\over 2}.$$
Now to find the term $P(D)$ in $(1)$, we first write
$$\tag{3}
P(D)=P(A\cap D)+P(B\cap D)+P(C\cap D)
$$
this is allowed since $A$, $B$, and $C$ are mutually exclusive events and one of the three must occur; as sets, $D$ can be written as the disjoint union $D= (A\cap D)\cup (B\cap D)\cup (C\cap D)$.
On to calculating the terms in $(3)$:
We have already calculated $P(C\cap D)$.
To find $P(A\cap D)$:
$$\tag{4}\textstyle
P(A\cap D)=P(A)P(D\mid A)={1\over3}\cdot{1\over4}={1\over12}.
$$
To find $P(B\cap D)$:
$$\tag{5}\textstyle
P(B\cap D)=P(B)P(D\mid B)={1\over6}\cdot{1\over8}={1\over48}.
$$
So, substituting the information from $(2)$, $(4)$ and $(5)$ into equation $(3)$, we have
$$\textstyle
P(D)= {1\over12}+{1\over48}+{1\over2} ={29\over 48}.
$$
Using this and $(1)$ and $(2)$ we finally obtain
$$
P(C\mid D) = {P(C\cap D)\over P(D)}={ 1/2\over 29/48}= {48\over 2\cdot 29}={24\over29}.
$$
A: 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}$
