In a competitive examination, an examinee either guesses or copies or knows the answer to amultiple choice question with four choices. The probability that he makes a guess is `1/3` and the probability that he copies the answer is 1/6. The probability that the answer is correct, given that he copiedit, is `1/8`. Find the probability that he knows the answer to the question, given that he correctly answered

Answer

B, D

Explanation

Let `E_(1),E_(2),E_(3)` and A be the evente defined as
`E_(1)=` the examinee copies the answer
`E_(2) =` the examinee copies the answer
`E_(3)=` the examine knows the answer
and A = the examinee answer correctly
We, have `P(E_(1))=(1)/(3),P(E_(2))=(1)/(6)`
Since, `E_(1),E_(2),E_(3)` are mutually exclusive and exhaustive events.
`therefore P(E_(1))+P(E_(2))+P(E_(3))=1`
`rArr P(E_(3) = 1-(1)/(3)-(1)/(6)=(1)/(2)`
If `E_(1)` has already occured, then the examine gueses. Since, there are four choices out of which only one is correct, therefore the probability that he answer correctly given that he has made a guess is 1/4.
i.e. `P(A//E_(1))=(1)/(4)`
It is given that, `P(A//E_(2))=(1)/(8)`
and `P(A//E_(3))` = probability that he answer correctly given that he know the answer = 1
By Baye's theorem, we have
`P(E_(3)//A)=(P(E_(3))*P(A//E_(3)))/([P(E_(1))*P(A//E_(1))+P(E_(2))*P(A//E_(2))+P(E_(3))*P(A//E_(3))])`
`therefore P(E_(3)//A)=((1)/(2)xx1)/(((1)/(3)xx(1)/(4))+((1)/(6)xx(1)/(8))+((1)/(2)xx1))=(24)/(29)`