Question

In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If he gets the correct answer to a question, then the probability that he was guessing, is

• A. 37/40
• B. 1/37
• C. 36/37
• D. 1/9

Answer 1

Answer: (B)

1/37

Answer 2

We define the following events :

A₁ : He knows the answer.

A₂: He does not know the answer

E : He gets the correct answer.

Then P(A₁) = 9/10, P(A₂) = 1 - $\frac{9}{10}\mspace{6mu} = \mspace{6mu}\frac{1}{10},\mspace{6mu} P\left( \frac{E}{A_{1}} \right) = 1$,

$P\left( \frac{E}{A_{2}} \right) = \frac{1}{4}$

∴ Required probability

= $P\left( \frac{A_{2}}{E} \right) = \frac{P(A_{2})P(E/A_{2})}{P(A_{1})P(E/A_{1}) + P(A_{2})P(E/A_{2})} = \frac{1}{37}$