Question

In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 13 and the probability that he copies the answer is 16. The probability that his answer is correct, given that he copies it, is 18. Find the probability that he knows the answer to the question, given that he correctly answered it.

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Solution

The correct option is **D** 2429

Let E1.E2,E3 and A be the events defined as follows :

E1 = the examinee guesses the answer

E2 = the examinee copies the answer

E3 = the examinee knows the answer and the examinee answers correctly

We have P(E1)=13,P(E2)=16

Since E1,E2 and E3 are mutually exclusive and exhaustive events therefore

P(E1)+P(E2)+P(E3)=1

⇒P(E3)=12

If E1 has already occurred, then the examinee guesses. Since

there are four choices out of which only one is correct, therefore the probability that he answers correctly given that he has made a guess is 14 i.e., P(AE1)=14

It is given that P(AE2)=18 and P(AE3) is the probability that he answers correctly given that he knew the answer = 1

By Baye's rule,

Required probability = P(E3A)

=P(E3)P(AE3)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)=12×113×14+16×18+12×1=2429

Let E1.E2,E3 and A be the events defined as follows :

E1 = the examinee guesses the answer

E2 = the examinee copies the answer

E3 = the examinee knows the answer and the examinee answers correctly

We have P(E1)=13,P(E2)=16

Since E1,E2 and E3 are mutually exclusive and exhaustive events therefore

P(E1)+P(E2)+P(E3)=1

⇒P(E3)=12

If E1 has already occurred, then the examinee guesses. Since

there are four choices out of which only one is correct, therefore the probability that he answers correctly given that he has made a guess is 14 i.e., P(AE1)=14

It is given that P(AE2)=18 and P(AE3) is the probability that he answers correctly given that he knew the answer = 1

By Baye's rule,

Required probability = P(E3A)

=P(E3)P(AE3)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)=12×113×14+16×18+12×1=2429

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