JEE Advanced 2024 — Paper 1 — Question 2

Consider A : student's answer is correct

$\mathrm{E}_{1}$ : He knows the answer.

$E_{2}$ : He guesses the answer.

$P\left(A / E_{2}\right)=\frac{1}{2}, P\left(A / E_{1}\right)=1$

Let $P\left(E_{1}\right)=1-x \Rightarrow P\left(E_{2}\right)=x$

Given, $P\left(E_{2} / A\right)=\frac{1}{6}=\frac{P\left(E_{2}\right) P\left(A / E_{2}\right)}{P\left(E_{1}\right) P\left(A / E_{1}\right)+P\left(E_{2}\right) P\left(A / E_{2}\right)}$

$\Rightarrow \frac{1}{6}=\frac{x \cdot \frac{1}{2}}{(1-x) \times 1+(x) \frac{1}{2}}=\frac{x}{2-x}$

$\Rightarrow x=\frac{2}{7}$

$\Rightarrow P($ req. $)=\frac{5}{7}$