Question

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with $4$ choices. 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 copied it, is $\frac{1}{8}$. Find the probability that he knew the answer to the question given that he correctly answered it.

• A. $\frac{11}{29}$
• B. $\frac{18}{29}$
• C. $\frac{15}{29}$
• D. $\frac{24}{29}$

Answer 1

The correct option is D $\frac{24}{29}$

Let $E_1,E_2,E_3$ and $A$ be the events defined as:
$E_1\to$ The examine guesses the answer.
$E_2\to$ The examine copies the answer
$E_3\to$ The examine knows the answer.
$A\to$ The examine answer correctly
We have, $P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{6}$
Since, $E_{1,}{E}_2,E_3$ are mutually exclusive and exhaustive events.
$\therefore P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1\Rightarrow P\left(E_3\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2}$
If $E_1$ has already occurred, then the examine gusses. Since, there are four choices out of which only one is correct, therefore the probability that he answer correctly given that he has made a guess is $\frac{1}{4}$
i.e $P\left(\frac{A}{E_1}\right)=\frac{1}{4}$
It is given that $P\left(\frac{A}{E_2}\right)=\frac{1}{8}$
and $P\left(\frac{A}{E_3}\right)=$ Probability that he answer correctly given that he knows the answer =$1$
By Baye's theorem, we have
$P\left(\frac{E_3}{A}\right)=\frac{P\left(E_3\right)P\left(\frac{A}{E_3}\right)}{[P\left(E_1\right).P\left(\frac{A}{E_1}\right)+P\left(E_2\right).P\left(\frac{A}{E_2}\right)+P\left(E_3\right).P\left(\frac{A}{E_3}\right)]}$
$\therefore P\left(\frac{E_3}{A}\right)=\frac{\frac{1}{2}\times 1}{(\frac{1}{3}\times \frac{1}{4})+(\frac{1}{6}\times \frac{1}{8})+(\frac{1}{2}\times 1)}=\frac{24}{29}$