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In an entrance examination with multiple choice questions, with each question having four options and a single correct answer, suppose that only $20\%$ candidates think they know the answer to one difficult question and only half of them know it correctly and the other half get it wrong. The remaining candidates pick one option out of the four randomly and tick the same. If a candidate has correctly answered the question, what is the (conditional) probability that she knew the answer?

Given that

P(Know)=.2 , P(Don’t know)=.8

P(Correct/Know)=.5 , P(Wrong/Know)=.5

P(Correct/ Don’t Know)=.25 , P(Wrong/ Don’t Know)=.75

We have to find P(Know/Correct)

By Bayes’ Theorem,

$P(Know/Correct)=\frac{P(Know \cap Correct )}{P(Correct)}$

$=\frac{P(Know)\times P(Correct/Know)}{P(Know\cap Correct)\dotplus P(Don't Know\cap Correct)}$

$=\frac{P(Know)\times P(Correct/Know)}{P(Know) \times P(Correct/Know)\dotplus P(Don't Know)\times P(Correct/(Don't Know)}$

$=\frac{.2 \times .5}{.2 \times .5 \dotplus .8 \times .25}$

$=\frac{.1}{.3}$

$=.333$

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