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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 the student knows the answer given that he answered it correctly?

$\begin{array}{1 1}12/13 \\ 1/13 \\ 6/13 \\ 8/13 \end{array} $

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  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Let $E_1$: student knows the answer $\rightarrow$ P($E_1$) = $\large\frac{3}{4}$
Let $E_2$: student guesses the answer $\rightarrow$ P($E_2$) = $\large\frac{1}{4}$
Probability that a student answered correctly knowing the answer = P (A|$E_1$) = 1.
Probability that a student answered by guessing = P (A|$E_2$) = $\large\frac{1}{4} $
We need to find the probability that a student who is chosen from random knows the answer given that he answered correctly:
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
Using Baye's theorem, P ($E_1$|A) =$ \Large\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{4}.\frac{1}{4}}$ = $\large\frac{3 \times 16 }{(12+1) \times 4} = \frac{12}{13}$
answered Jun 19, 2013 by balaji.thirumalai

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