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A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices.

As the teacher you want the test to measure what the student knows. If the student answers a question correctly what’s the probability they knew the answer?

I am getting 0.787 I.e. 78.7 % but answer is 77.4 %. Can someone verify This problem will be solved using Bayes Theorem.

Let C be the event that the student answered corectly

P(C) = $\frac{1}{2}+\frac{1}{4}*\frac{1}{3}+\frac{1}{4}*\frac{1}{4} = \frac{31}{48}$

Let X be the event that student knew the answer.

Then , P(X/C) = $\frac{P(X)*P(C/X)}{P(C)} = \frac{\frac{1}{2}*1}{\frac{31}{48}} = \frac{48}{62} = 0.7741$.

Thus in percentage = 77.41%

i feel language isn't clear in the first go..
what did you not understand?
its clear now .

that time didnt read the question properly