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

Accordingly, $P ( A $ and $B)=0.5$,  $P ($ not $A$ and  $B )=0.1,$

i.e., $P \left( A^{\prime} \cap B ^{\prime}\right)=0.1$

$P ( A )=0.75$

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

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

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

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

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

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

$\Rightarrow P(B)=0.65$

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