Let $K$ be the event that the student knows the answer.
Let $E$ be the event that the student can eliminate one choice.
Let $G$ be the event that the student guesses from all 4 choices.
Let $C$ be the event that the student answers correctly.
From the problem description, we have the following probabilities:
- $P(K)=0.5=\dfrac{1}{2}$
- $P(E)=0.25=\dfrac{1}{4}$
- $P(G)=1-P(K)-P(E)=1-0.5-0.25=0.25=\dfrac{1}{4}$
The conditional probabilities of getting the correct answer given their state of knowledge are:
- $P(C\mid K)=1$ (If they know the answer, they get it right)
- $P(C\mid E)=\dfrac{1}{3}$ (Guessing from the $3$ remaining choices)
- $P(C\mid G)=\dfrac{1}{4}$ (Guessing from all $4$ choices)
We want to find $P(K\mid C)$, the probability that the student knows the answer given that they answered correctly. We use Bayes’ Theorem:
\[
P(K\mid C)=\frac{P(C\mid K)P(K)}{P(C)}
\]
First, calculate the total probability of a correct answer, $P(C)$, using the Law of Total Probability:
\[
P(C)=P(C\mid K)P(K)+P(C\mid E)P(E)+P(C\mid G)P(G)
\]
\[
P(C)=\left(1\cdot \frac{1}{2}\right)+\left(\frac{1}{3}\cdot \frac{1}{4}\right)+\left(\frac{1}{4}\cdot \frac{1}{4}\right)
\]
\[
P(C)=\frac{1}{2}+\frac{1}{12}+\frac{1}{16}
\]
To add these fractions, find the least common multiple of $2, 12,$ and $16,$ which is $48:$
\[
P(C)=\frac{24}{48}+\frac{4}{48}+\frac{3}{48}=\frac{31}{48}
\]
Now, substitute this back into our Bayes’ Theorem equation:
\[
P(K\mid C)=\frac{1\cdot \frac{1}{2}}{\frac{31}{48}}
=\frac{\frac{1}{2}}{\frac{31}{48}}
=\frac{1}{2}\cdot \frac{48}{31}
=\frac{24}{31}
\]
The correct option is A.
