A student answers a multiple-choice examination question that offers four possib

Question

A student answers a multiple-choice examination question that offers four possible answers. Suppose that the probability that the student knows the answer to the question is 0.8 and the probability that the student will guess is 0.2. Assume that if the student guesses, the probability of selecting the correct answer is 0.25. If the student correctly answers a question, what is the probability that student really knew the correct answer?

I have set G = student guesses the answer, Gc = student knows the answer, and C = student chooses the correct answer. I know that I have to find P(Gc|C) where P(Gc|C) = [P(C|Gc)*P(Gc)]/[P(C|Gc)*P(Gc) + P(C|G)*P(G)].
I know that P(Gc) = .8, P(G) = .2, and P(C) = .25 but I do not know what P(C|Gc) is or how to get it. Could someone help me?
I thank God for all of your help, ~beauty
I have set G = student guesses the answer, Gc = student knows the answer, and C = student chooses the correct answer. I know that I have to find P(Gc|C) where P(Gc|C) = [P(C|Gc)*P(Gc)]/[P(C|Gc)*P(Gc) + P(C|G)*P(G)].
I know that P(Gc) = .8, P(G) = .2, and P(C) = .25 but I do not know what P(C|Gc) is or how to get it. Could someone help me?
I thank God for all of your help, ~beauty

Explanation / Answer

If the student knows the answere, then p(C|Gc)=1. Now, P(Gc|c)=P(c|Gc).P(Gc)/[P(c|Gc).P(Gc)+P(c|G).P(G)] =1(0.8)/[1(0.8)+0.25(0.2) =0.8/0.85 =16/17 =0.941176.

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