NIMCET 2016 Question Paper with solutions

Total number of possible outcomes = 10
$\therefore \mathit{P}\left(\mathit{A}\right)=\frac{4}{10}=\frac{2}{5}$
Assume that 8 has n outcomes. Then,
Given that, A and B are independent events.
$\therefore \mathit{P}\left(\mathit{A}\cap \mathit{B}\right)=\mathit{P}\left(\mathit{A}\right)\cdot \mathit{P}\left(\mathit{B}\right)$
$\Rightarrow \mathit{P}\left(\mathit{A}\cap \mathit{B}\right)=\frac{2}{5}\times \frac{\mathit{n}}{10}$
$\Rightarrow \mathit{P}\left(\mathit{A}\cap \mathit{B}\right)=\frac{\frac{2\mathit{n}}{5}}{10}$
$\because \frac{2\mathit{n}}{5}$ is an integer which is lies between 0 and 10 and $\frac{\mathit{n}}{5}$ is an integers which is lies between 0 and 5.
$\therefore \mathit{n}=5,10$