Q. In a test, an examines either guesses or copies or knows the answer to a multiple choice questions with four choices. The probability that he makes a guess is $\frac{1}{3}$and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct given that he copies it is $\frac{1}{8}$. The probability that his answer is correct, given that he guesses it is $\frac{1}{4}$.The probability that they knew the answer to the questions given that he correctly answered is
Explanation
(c)
Let $E_{1}{,\ E}_{2\ }$and $E_{3}$ are the examines guesses, copies and knows the answer
and $E =$Event that he answers correctly
Then, $P\left( E_{1} \right) = \frac{1}{3},P\left( E_{2} \right) = \frac{1}{6}$
and $P\left( E_{3} \right) = 1 - \left( \frac{1}{3} + \frac{1}{6} \right) = \frac{1}{2}$
$\therefore$ Required probability$= P\left( \frac{E_{3}}{E} \right)$
$= \frac{P\left( \frac{E}{E_{3}} \right).P(E_{3})}{P\left( \frac{E}{E_{1}} \right).P\left( E_{1} \right) + P\left( \frac{E}{E_{2}} \right).P\left( E_{2} \right) + P\left( \frac{E}{E_{3}} \right).P(E_{3})}$
$= \frac{1 \times \frac{1}{2}}{\left( \frac{1}{4} \times \frac{1}{3} \right) + \left( \frac{1}{8} \times \frac{1}{6} \right) + \left( 1 \times \frac{1}{2} \right)} = \frac{24}{29}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.