Question

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

• A. $\frac{17}{39}$
• B. $\frac{13}{29}$
• C. $\frac{24}{29}$
• D. $\frac{24}{39}$

Answer 1

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

Let $E_1.E_2,E_3$ and A be the events defined as follows :
$E_1$ = the examinee guesses the answer
$E_2$ = the examinee copies the answer
$E_3$ = the examinee knows the answer and the examinee answers correctly
We have $P\left(E_1\right)=\frac{1}{3},P\left(E_2\right)=\frac{1}{6}$
Since $E_1,E_2$ and $E_3$ are mutually exclusive and exhaustive events therefore
$P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1$
$\Rightarrow P\left(E_3\right)=\frac{1}{2}$
If $E_1$ has already occurred, then the examinee guesses. Since
there are four choices out of which only one is correct, therefore the probability that he answers 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 $P\left(\frac{A}{E_3}\right)$ is the probability that he answers correctly given that he knew the answer = 1
By Baye's rule,
Required probability = $P\left(\frac{E_3}{A}\right)$
$=\frac{P\left(E_3\right)P\left(\frac{A}{E_3}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+P\left(E_3\right)P\left(\frac{A}{E_3}\right)}=\frac{\frac{1}{2}\times 1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}=\frac{24}{29}$