In answering a question on a multiple choice test a student eithe… - Vidyadip

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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Answer

Let A, $E_1$ and $E_2$ denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{3}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=1$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{1}{4}$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{4}\times1}{\frac{3}{4}\times1+\frac{1}{4}\times\frac{1}{4}}$
$=\frac{3}{3+\frac{1}{4}}=\frac{12}{13}$

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