Question

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 weak student knows only $20\%$ of the correct answers. If a weak student gets the correct answer, the probability that he was guessing is

• A. 0.03
• B. 0.27
• C. 0.40
• D. 0.50

Answer 1

Answer: (D)

Let $E_1$ be the event that a weak student know the answer and $E_2$ 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 $=P\left(\frac{E_2}{A}\right)$
Clearly, $P\left(E_1\right)=0.2,P\left(E_2\right)=0.8$
$P\left(\frac{A}{E_1}\right)=1$
$\Rightarrow P\left(\frac{A}{E_2}\right)=0.25$
$\therefore P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right)\cdot P\left(A/E_2\right)}{P\left(E_1\right)\cdot P\left(A/E_1\right)+P\left(E_2\right)\cdot P\left(A/E_2\right)}$
$=\frac{0.8\times 0.25}{0.2\times 1+0.8\times 0.25}$
$=\frac{0.2}{0.2+0.2}=\frac{0.2}{0.4}=\frac{1}{2}=0.5$