Question

In a test, an examinee either guess or copies or knows the answer to a multiple choice question with four choices and only one correct option. The probability that he makes a guess is $\frac{1}{3}$. The probability that he copies the answer is $\frac{1}{6}$. The probability that the answer is correct, given that he copied it, is $\frac{1}{8}$. Find the probability that he knows the answer to the question, given that he correctly answered it.

• A. $\frac{23}{29}$
• B. $\frac{27}{29}$
• C. $\frac{24}{29}$
• D. $\frac{25}{29}$

Answer 1

Answer: (C)

Let $E_1$, $E_2$, $E_3$ and $A$ be the events defined as follows :
$E_1 =$ the examinee guess the answer,
$E_2 =$ the examinee copies the answer,
$E_3 =$ the examinee knows the answer and
$A =$ the examinee has answered the question correctly.
$P(E_1) = \frac{1}{3}$, $P(E_2) = \frac{1}{6}$ (given)
As $E_1$, $E_2$ and $E_3$ are mutually exclusive and exhaustive events,
$P(E_1) + P(E_2) + P(E_3) = 1$
$\Rightarrow P(E_3) = 1- P ( E_1) - P (E_2) = 1 -\frac{1}{3}-\frac{1}{6} = \frac{1}{2}$
When $E_1$ has occurred, then the examinee guesses. Since, there are four choices and only one is correct, the probability that he answers correctly given that he has made a guess is $\frac{1}{4}$ i.e. $P(A|E_1) = \frac{1}{4}$
$P\left(A|E_{2}\right) = $ probability that he answers correctly given that he has copied $= \frac{1}{8}$
When $E_3$ has occurred i. e. the examinee knows the answer, then the probability that he answers correctly given that he knows the answer is $1$ (sure event) i.e. $P(A|E_3) = 1$.
We want to find $P(E_3|A)$
By Bayes' theorem, we have

$P\left(E_{3}|A\right) = \frac{P\left(E_{3}\right)P\left(A |E_{3}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)+ P\left(E_{3}\right)P\left(A|E_{3}\right)}$
$= \frac{\frac{1}{2}\cdot1}{\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{6}\cdot\frac{1}{8}+\frac{1}{2}\cdot1} = \frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}$
$= \frac{1}{2}\times\frac{48}{29} = \frac{24}{29}$