First slide

Theorems of probability

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 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

Moderate

A

$\frac{24}{29}$
B

$\frac{1}{4}$
C

$\frac{3}{4}$
D

$\frac{1}{2}$

Solution

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.
$∴ P (E_{1}) + P (E_{2}) + P (E_{3}) = 1$
$∴ P (E_{3}) = 1 - P (E_{1}) - P (E_{2}) = 1 - \frac{1}{3} - \frac{1}{6} = \frac{1}{2}$
Now, $P (\frac{E}{E_{1}}) =$ probability of getting correct answer by guessing = $\frac{1}{4}$ (since 4 alternatives)
$P (\frac{E}{E_{2}}) =$ probability of answering correctly by copying $= \frac{1}{8}$
And $P (\frac{E}{E_{3}}) =$ probability of answering correctly by knowing = 1
Clearly, $(\frac{E_{3}}{E})$ is the event he knew the answer to the question that he correctly answered it. Using Baye’s theorem $P (\frac{E_{3}}{E})$ = $\frac{P (E_{3}) . P (\frac{E}{E_{3}})}{P (E_{1}) . P (\frac{E}{E_{1}}) + P (E_{2}) . P (\frac{E}{E_{2}}) + P (E_{3}) . P (\frac{E}{E_{3}})}$
$= \frac{\frac{1}{2} \times 1}{\frac{1}{3} \times \frac{1}{4} + \frac{1}{6} \times \frac{1}{8} + \frac{1}{2} \times 1} = \frac{24}{29} (b y B a y e s' t h e o r e m)$

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