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}$.

Similar Questions

$A$ bag contains $(N+1)$ coins: $N$ fair coins, and one coin with 'Head' on both sides. $A$ coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then $N$ is equal to:

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