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In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?
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Verified Answer
Let the event $\mathrm{E}_1=$ student knows the answer,
$\mathrm{E}_2=$ He gusses the answer
$$
\mathrm{P}\left(\mathrm{E}_1\right)=\frac{3}{4}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{4}
$$
Let $\mathrm{A}$ is the event that answer is correct, if the student knows the answer
$\begin{array}{lll}\Rightarrow \text { Answer is correct } & \therefore & \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=1 \\ \text { If he guesses the answer } & \therefore & \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{4}\end{array}$
$\therefore \quad$ Probability that a student knows the answer given that answer is correct is,
$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$
$$
\begin{aligned}
&=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)} \\
&=\frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1+\frac{1}{4} \times \frac{1}{4}}=\frac{\frac{3}{13}}{\frac{13}{16}}=\frac{3}{4} \times \frac{16}{13}=\frac{12}{13}
\end{aligned}
$$
$\mathrm{E}_2=$ He gusses the answer
$$
\mathrm{P}\left(\mathrm{E}_1\right)=\frac{3}{4}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{4}
$$
Let $\mathrm{A}$ is the event that answer is correct, if the student knows the answer
$\begin{array}{lll}\Rightarrow \text { Answer is correct } & \therefore & \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=1 \\ \text { If he guesses the answer } & \therefore & \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{4}\end{array}$
$\therefore \quad$ Probability that a student knows the answer given that answer is correct is,
$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$
$$
\begin{aligned}
&=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)} \\
&=\frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1+\frac{1}{4} \times \frac{1}{4}}=\frac{\frac{3}{13}}{\frac{13}{16}}=\frac{3}{4} \times \frac{16}{13}=\frac{12}{13}
\end{aligned}
$$
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