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)} $
$= \frac{\frac{1}{2}\times1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}= \frac{24}{29}$