In answering a question on a multiple choice | vedclass

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Question

(A) Let $E_{1}$ and $E_{2}$ be the respective events that the student knows the answer and he guesses the answer.Let $A$ be the event that the answer

Vedclass · Accepted Answer

$\frac{12}{13}$

Vedclass · Answer

(A) Let $E_{1}$ and $E_{2}$ be the respective events that the student knows the answer and he guesses the answer.Let $A$ be the event that the answer is correct.Given probabilities are $P(E_{1}) = \frac{3}{4}$ and $P(E_{2}) = \frac{1}{4}$.The probability that the student answers correctly given that he knows the answer is $P(A|E_{1}) = 1$.The probability that the student answers correctly given that he guesses is $P(A|E_{2}) = \frac{1}{4}$.We need to find $P(E_{1}|A)$. By Bayes' theorem:$P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$$P(E_{1}|A) = \frac{\frac{3}{4} 	imes 1}{(\frac{3}{4} 	imes 1) + (\frac{1}{4} 	imes \frac{1}{4})}$$P(E_{1}|A) = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} = \frac{\frac{3}{4}}{\frac{12+1}{16}} = \frac{\frac{3}{4}}{\frac{13}{16}}$$P(E_{1}|A) = \frac{3}{4} 	imes \frac{16}{13} = \frac{12}{13}$.

Box	Black	White	Red	Blue
$I$	$3$	$4$	$5$	$6$
$II$	$2$	$2$	$2$	$2$
$III$	$1$	$2$	$3$	$1$
$IV$	$4$	$3$	$1$	$5$

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