The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1.

Let A and B be the events of passing English and Hindi examinations respectively.

Accordingly, $P(A$ and $B)=0.5, P($ not $A$ and not $B)=0.1$, i.e., $P\left(A^{\prime} \cap B^{\prime}\right)=0.1$

P(A) = 0.75

Now, $(A \cup B)^{\prime}=\left(A^{\prime} \cap B^{\prime}\right) \quad[$ De Morgan's law $]$

$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=0.1$

$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}=1-0.1=0.9$

We know that P(A or B) = P(A) + P(B) – P(A and B)

$\therefore 0.9=0.75+\mathrm{P}(\mathrm{B})-0.5$

$\Rightarrow P(B)=0.9-0.75+0.5$

$\Rightarrow P(B)=0.65$

Thus, the probability of passing the Hindi examination is 0.65.