There are 3 True or False questions in an exam, if a candidate knows the answer she/he answers it correctly, otherwise a guess is made and the probability of getting it right is 1/2, an examiner assumes that every candidate knows no answer, 1 answer, 2 answers, 3 answers with equal probabilities, a candidate answered two of the three questions correctly, what is the probability that this candidate knew the answer to only one of them?
(A) 1/11. (B) 2/11. (C) 3/11. (D) 4/11.
So for the required probability the numerator must have the condition that out of 2 guesses 1 is correct and another is not so the probability can be $3C2$ × $1/2$ × $1/2$ but what will be the denominator?
We perform a more detailed computation, finding the probability that the student got $y=2$ right and knew $x$ for $x=0,1,2$: $$P(y=2\mid x=0)=\binom32(1/2)^3=\frac38$$ $$P(y=2\mid x=1)=\binom21(1/2)^2=\frac12$$ $$P(y=2\mid x=2)=\frac12$$ (We don't need to calculate for $x=3$ since in that case we would have $y=3)$. Finally, by total probability and since $x=0,1,2,3$ with equal probability, $$P(x=1\mid y=2)=\frac{1/2}{3/8+1/2+1/2}=\frac4{11}$$