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

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(B) Let the events be defined as follows:$A_1$: The student knows the answer.$A_2$: The student does not know the answer (he is guessing).$E$: The stu

Vedclass · Accepted Answer

$\frac{1}{37}$

Vedclass · Answer

(B) Let the events be defined as follows:$A_1$: The student knows the answer.$A_2$: The student does not know the answer (he is guessing).$E$: The student gets the correct answer.Given:$P(A_1) = 0.9 = \frac{9}{10}$$P(A_2) = 1 - 0.9 = 0.1 = \frac{1}{10}$$P(E|A_1) = 1$ (If he knows the answer,he will definitely get it correct).$P(E|A_2) = \frac{1}{4}$ (If he is guessing,there is a $1$ in $4$ chance of being correct).We need to find the probability that he was guessing given that he got the correct answer,i.e.,$P(A_2|E)$.Using Bayes' Theorem:$P(A_2|E) = \frac{P(A_2)P(E|A_2)}{P(A_1)P(E|A_1) + P(A_2)P(E|A_2)}$$P(A_2|E) = \frac{(\frac{1}{10}) 	imes (\frac{1}{4})}{(\frac{9}{10}) 	imes (1) + (\frac{1}{10}) 	imes (\frac{1}{4})}$$P(A_2|E) = \frac{\frac{1}{40}}{\frac{9}{10} + \frac{1}{40}} = \frac{\frac{1}{40}}{\frac{36+1}{40}} = \frac{1}{37}$.

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 $90\%$. If he gets the correct answer to a question,then the probability that he was guessing is:

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