Let the events A, B, C are the sections in which students pass.

$\Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\mathrm{p}, \mathrm{P}\left(\mathrm{B}\right)=\mathrm{q}, \mathrm{P}\left(\mathrm{C}\right)=\frac{1}{2}$

The student will be successful if he able to pass the section A and either of section B or Section C = (Probability of passing in A) \times (Probability of selecting either of section B or C \times (Probability of passing B + Probability of passing C)) $=\mathrm{P}\left(\mathrm{A}\right)\times \left(\frac{1}{2}\times \left(\mathrm{P}\right(\mathrm{B})+\mathrm{P}(\mathrm{C}\left)\right)=\mathrm{p}\times \left(\frac{1}{2}\times \left(\mathrm{q}+\frac{1}{2}\right)\right)=\frac{\mathrm{pq}}{2}+\frac{\mathrm{p}}{4}\right.$

But Probability that student is successful in the exam $=\frac{1}{2}$

$\Rightarrow \frac{1}{2}=\frac{\mathrm{pq}}{2}+\frac{\mathrm{p}}{4}$

⇒ 2pq + p = 2

From the given option only Option 4 i.e., p = 1, $\mathrm{q}=\frac{1}{2}$ satisfies the equation.

Hence, options 1, 2, 3 are wrong.