Mathematics - Probability Question with Solution | TestHub

Given,

Two fair dices are rolled,

So total sample space will be $\mathrm{n}\left(\mathrm{s}\right)=36$

Now given: $\mathrm{N}-2, \sqrt{3 \mathrm{N}}, \mathrm{N}+2$ are in G.P.

So, $3\mathrm{N}=\left(\mathrm{N}-2\right)\left(\mathrm{N}+2\right)$

$\Rightarrow 3\mathrm{N}={\mathrm{N}}^2-4$

$\Rightarrow {\mathrm{N}}^2-3\mathrm{N}-4=0$

$\Rightarrow \left(\mathrm{N}-4\right)\left(\mathrm{N}+1\right)=0$

$\Rightarrow \mathrm{N}=4$  or $\mathrm{N}=-1$ (rejected)

Now favourable cases for sum $4 \equiv \left\{\left(1,3\right),\left(3,1\right),\left(2,2\right)\right\}$

So, $n\left(A\right)=3$

Hence, $P\left(A\right)=\frac{3}{36}=\frac{1}{12}=\frac{4}{48}\Rightarrow k=4$