Question

In a test, an examinee either guesses or copies or knows that answer to a multiple-choice question which has four choices. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies is $\frac{1}{6}$. The probability that his answer is correct, given he copied it, is $\frac{1}{8}$. Find the probability that he knew the answer to the question, given that he answered it correctly.

Answer 1

Step 1: Find the probability that the examinee knows the answer.

Since, it is given that, the probability $P\left(G\right)$ that the examinee guesses the answer is $\frac{1}{3}$.

Also, it is given that, the probability $P\left(C\right)$ that the examinee copies the answer is $\frac{1}{6}$.

We know that the sum of probabilities is always equal to $1$.

Therefore, the probability $P\left(K\right)$ that the examinee knows the answer is given by:

$$\begin{gathered} \Rightarrow P\left(G\right)+P\left(C\right)+P\left(K\right)=1 \\ \Rightarrow P\left(K\right)=1-P\left(G\right)-P\left(C\right) \\ \Rightarrow P\left(K\right)=1-\frac{1}{3}-\frac{1}{6} \\ \Rightarrow P\left(K\right)=\frac{1}{2} \end{gathered}$$

So, the probability $P\left(K\right)$ that the examinee knows the answer is $\frac{1}{2}$.

Step 2: Find the required probability.

Assume that, $R$ is an event that the examinee's answer is correct.

So, the probability $P\left(\frac{R}{G}\right)$ that the examinee's answer is correct by guessing.

$P\left(\frac{R}{G}\right)=\frac{1}{4}$ {Because the multiple-choice question has four choices}

The probability $P\left(\frac{R}{K}\right)$ that the examinee's answer is correct when he knows the answer is $1$ because it is a sure event.

Since it is given that the probability $P\left(\frac{R}{C}\right)$that the examinee's answer is correct by copying is $\frac{1}{8}$.

Now, apply Baye's theorem to find the probability $P\left(\frac{K}{R}\right)$ that the examinee knew the answer given that his answer is correct.

$$\begin{gathered} \Rightarrow P\left(\frac{K}{R}\right)=\frac{P\left({\displaystyle \frac{R}{K}}\right)·P\left(K\right)}{P\left(\frac{R}{G}\right)·P\left(G\right)+P\left(\frac{R}{C}\right)·P\left(C\right)+P\left(\frac{R}{K}\right)·P\left(K\right)} \\ \Rightarrow P\left(\frac{K}{R}\right)=\frac{1·{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{4}}·{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{8}}·{\displaystyle \frac{1}{6}}+1·{\displaystyle \frac{1}{2}}} \\ \Rightarrow P\left(\frac{K}{R}\right)=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{12}+\frac{1}{48}+\frac{1}{2}}} \\ \Rightarrow P\left(\frac{K}{R}\right)=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{4+1+24}{48}}} \\ \Rightarrow P\left(\frac{K}{R}\right)=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{29}{48}}} \\ \Rightarrow P\left(\frac{K}{R}\right)=\frac{24}{29} \end{gathered}$$

Hence, the probability that the examinee knew the answer to the question, given that he answered it correctly is $\frac{24}{29}$.