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In a certain recruitment test with multiple choice questions, there are four options to each question, out of which only one is correct. An intelligent student knows $90 \%$ of the correct answers while a weals student knows only $20 \%$ of the correct answers. If a weals student gets the correct answer, the probability that he was guessing is
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Verified Answer
The correct answer is:
0.50
Let $E_1$ be the event that a weak student know the answer and $B_2$ be the event that a weak student guess the answer and $A$ be the event that a weak student gets the correct answer.
Then, required probability $=P\left(\frac{B_2}{A}\right)$
Clearly, $P\left(B_1\right)=0.2, P\left(B_2\right)=0.8$
$$
\begin{aligned}
& P\left(\frac{A}{B_1}\right)=1 \Rightarrow P\left(\frac{A}{B_2}\right)=0.25 \\
& \therefore P\left(\frac{B_2}{A}\right)=\frac{P\left(B_2\right) \cdot P\left(A / B_2\right)}{P\left(B_1\right) \cdot P\left(A / B_1\right)+P\left(B_2\right) \cdot P\left(A / B_2\right)} \\
& =\frac{0.8 \times 0.25}{0.2 \times 1+0.8 \times 0.25} \\
& =\frac{0.2}{0.2+0.2}=\frac{0.2}{0.4}=\frac{1}{2}=0.5 \\
&
\end{aligned}
$$
Then, required probability $=P\left(\frac{B_2}{A}\right)$
Clearly, $P\left(B_1\right)=0.2, P\left(B_2\right)=0.8$
$$
\begin{aligned}
& P\left(\frac{A}{B_1}\right)=1 \Rightarrow P\left(\frac{A}{B_2}\right)=0.25 \\
& \therefore P\left(\frac{B_2}{A}\right)=\frac{P\left(B_2\right) \cdot P\left(A / B_2\right)}{P\left(B_1\right) \cdot P\left(A / B_1\right)+P\left(B_2\right) \cdot P\left(A / B_2\right)} \\
& =\frac{0.8 \times 0.25}{0.2 \times 1+0.8 \times 0.25} \\
& =\frac{0.2}{0.2+0.2}=\frac{0.2}{0.4}=\frac{1}{2}=0.5 \\
&
\end{aligned}
$$
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