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, $P\left(E_{1}\right)=0.2,\, P\left(E_{2}\right)=0.8$
$P\left(\frac{A}{E_{1}}\right)=1$
$\Rightarrow P\left(\frac{A}{E_{2}}\right)=0.25$
$\therefore P\left(\frac{E_{2}}{A}\right) =\frac{P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}{P\left(E_{1}\right) \cdot P\left(A / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A / E_{2}\right)}$
$= \frac{0.8 \times 0.25}{0.2 \times 1+0.8 \times 0.25}$
$=\frac{0.2}{0.2+0.2}=\frac{0.2}{0.4}=\frac{1}{2}=0.5$