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Question

# In answering a question on a multiple choice test a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that a student knows the answer given that he answered it correctly?

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Solution

## Let A, E1 and E2 denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively. $\therefore P\left({E}_{1}\right)=\frac{3}{4}\phantom{\rule{0ex}{0ex}}P\left({E}_{2}\right)=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}P\left(A/{E}_{1}\right)=1\phantom{\rule{0ex}{0ex}}P\left(A/{E}_{2}\right)=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\mathrm{Required}\mathrm{probability}=P\left({E}_{1}/A\right)=\frac{P\left({E}_{1}\right)P\left(A/{E}_{1}\right)}{P\left({E}_{1}\right)P\left(A/{E}_{1}\right)+P\left({E}_{2}\right)P\left(A/{E}_{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\frac{3}{4}×1}{\frac{3}{4}×1+\frac{1}{4}×\frac{1}{4}}\phantom{\rule{0ex}{0ex}}=\frac{3}{3+\frac{1}{4}}=\frac{12}{13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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