In an entrance test,there are multiple-choice | vedclass

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Question

(B) Let $E_1$ be the event that the student knows the answer and $E_2$ be the event that the student guesses the answer. Given $P(E_1) = 9/10$,then $P

Vedclass · Accepted Answer

$\frac{1}{37}$

Vedclass · Answer

(B) Let $E_1$ be the event that the student knows the answer and $E_2$ be the event that the student guesses the answer. Given $P(E_1) = 9/10$,then $P(E_2) = 1 - 9/10 = 1/10$. Let $E$ be the event that the answer is correct. If the student knows the answer,the probability of answering correctly is $P(E|E_1) = 1$. If the student guesses,since there are $4$ options and $1$ is correct,the probability of answering correctly is $P(E|E_2) = 1/4$. Using Bayes' theorem,the probability that the student was guessing given that the answer is correct is $P(E_2|E) = \frac{P(E|E_2)P(E_2)}{P(E|E_1)P(E_1) + P(E|E_2)P(E_2)}$. Substituting the values: $P(E_2|E) = \frac{(1/4) 	imes (1/10)}{(1) 	imes (9/10) + (1/4) 	imes (1/10)} = \frac{1/40}{9/10 + 1/40} = \frac{1/40}{36/40 + 1/40} = \frac{1/40}{37/40} = \frac{1}{37}$.

	Red	Blue	Green
Bag $I$	$3$	$2$	$5$
Bag $II$	$4$	$3$	$3$
Bag $III$	$5$	$1$	$4$

In an entrance test,there are multiple-choice questions. There are four possible answers to each question,of which one is correct. The probability that a student knows the answer to a question is $9/10$. If he gets the correct answer to a question,then the probability that he was guessing is:

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