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 copied it is $\frac{1}{8}.$ Find the probability that he knew the answer to the question, given that he answered it correctly.

• A. $\frac{17}{21}$
• B. $\frac{23}{27}$
• C. $\frac{24}{29}$
• D. $\frac{21}{23}$

Answer 1

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

Let $E_1,E_2,E_3$ and A be the events defined as
$E_1$ = the examinee guesses the answer
$E_2$ = the examinee copies the answer
$E_3$ = the examinee knows the answer
and A = the examinee answers correctly
Wa 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 the examinee guesses, since there are four choices out of which only one is correct, the probability that he answers correctly is $\frac{1}{4}$
$\Rightarrow P\left(A|E_1\right)=\frac{1}{4}$
$Giventhat,P\left(A|E_2\right)=\frac{1}{8}$
$P\left(A|E_3\right)=1$
By Bayes' theorem, we have
$P\left(E_3|A\right)=\frac{P\left(E_3\right).P\left(A|E_3\right)}{\left[P\right(E_1).P(A|E_1)+P(E_2).P(A|E_2)+P(E_3).P(A|E_3\left)\right]}$
$\therefore P\left(E_3|A\right)=\frac{\frac{1}{2}\times 1}{(\frac{1}{3}\times \frac{1}{4})+(\frac{1}{6}\times \frac{1}{8})+(\frac{1}{2}\times 1)}=\frac{24}{29}$