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?
Best Answer
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 $$ P(C\mid D) = {P(C\cap D)\over P(D)}={ 1/2\over 29/48}= {48\over 2\cdot 29}={24\over29}. $$