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 $\frac{1}{3}$ . The probability that he copies is $\frac{1}{6}$ and the probability that his answer is correct given that he copied it is $\frac{1}{8}$ . The probability that he knew the answer to the question given that he correctly answered it, is

• A. $\frac{24}{29}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

Let $E_1$ be the event that the answer is guessed, $E_2$ be the event that the answer is copied, $E_3$ be the event that the examinee knows the answer and E be the event that the examinee answers correctly.
Given $P(E_1)=\frac{1}{3}.P(E_2)=\frac{1}{6}$
Assume that events $E_1,E_2$ & $E_3$ are exhaustive.
$\therefore P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1$
$\therefore P\left(E_3\right)=1-P\left(E_1\right)-P\left(E_2\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2}.$
Now, $P\left(\frac{E}{E_1}\right)\equiv$ Probability of getting correct answer by guessing $=\frac{1}{4}$ (Since 4 alternatives)
$P\left(\frac{E}{E_2}\right)\equiv$ Probability of answering correctly by
copying $=\frac{1}{8}$ and $P\left(\frac{E}{E_3}\right)\equiv$ Probability of answering correctly by knowing = 1
Clearly, $\left(\frac{E_3}{E}\right)$ is the event he knew the answer to the question given that he correctly answered
it. Using Baye’s theorem $P\left(\frac{E_3}{E}\right)$
$=\frac{P\left(E_3\right).P\left(\frac{E}{E_3}\right)}{P\left(E_1\right).P\left(\frac{E}{E_1}\right)+P\left(E_2\right).P\left(\frac{E}{E_2}\right)+P\left(E_3\right).P\left(\frac{E}{E_3}\right)}$
$=\frac{\frac{1}{2}\times 1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}=\frac{24}{29}$