Question

In a test, a candidate, either guesses or copy or knows the answer of a MCQ question with four choices.
The probability that he makes a guess is 1/3 & the probability that he copies the answer is 1/6. The
probability that his answer is correct, given that he copied it is 1/8. Find the probability that he knows
the answer to the question, given that he correctlyanswered it.

Pls give correct answer..​

Answer 1

Given

To find the probability that he knows the answer to the question (given that he correctly answered).

Let, P(A) is the probability of guessing = 1/3.

      P(B) is the probability of copying = 1/6.

      P(C) is the probability of knowing = 1 - (1/3 + 1/6) = 1 - 1/3 - 1/6

                                                               = 1/2

      P(D) is the probability that answer is correct

Using Baye's theorem,

      P(C/D) = [ $\frac{(P(D/C) . P(C))}{ (P(D/B) . P(B) + P(D/C) . P(C) + P(D/A) . P(A))}$ ]          

                   = [$\frac{(1.1/2)}{ ((1/8.1/6)+(1.1/2)+(1/4.1/3) }$]

       P(C/D) = 24 / 29

Probability that he knows the answer to the question is 24/29.

To learn more...

brainly.in/question/2256594  

Answer 2

Answer:

Given

To find the probability that he knows the answer to the question (given that he correctly answered).

Let, P(A) is the probability of guessing = 1/3.

P(B) is the probability of copying = 1/6.

P(C) is the probability of knowing = 1 - (1/3 + 1/6) = 1 - 1/3 - 1/6

= 1/2

P(D) is the probability that answer is correct

Using Baye's theorem,

P(C/D) = [ \frac{(P(D/C) . P(C))}{ (P(D/B) . P(B) + P(D/C) . P(C) + P(D/A) . P(A))}

(P(D/B).P(B)+P(D/C).P(C)+P(D/A).P(A))

(P(D/C).P(C))

]

= [\frac{(1.1/2)}{ ((1/8.1/6)+(1.1/2)+(1/4.1/3) }

((1/8.1/6)+(1.1/2)+(1/4.1/3)

(1.1/2)

P(C/D) = 24 / 29

Probability that he knows the answer to the question is 24/29.