NIMCET 2021 Question Paper with solutions

Three dices thrown,
Total outcomes when three dice drawn is $=6\times 6\times 6=216$
Total outcomes of triplets is
$\left(1,1,1\right),\left(2,2,2\right),\left(3,3,3\right),\left(4,4,4\right),\left(5,5,5\right),$ $\left(6,6,6\right)$
Probability of triplet when three dice thrown once is $=\frac{6}{216}=\frac{1}{36}$
Required probability on throwing three throws of three dice not more than twice,
$\Rightarrow \mathit{P}\left(\mathit{x}=0\right)+\mathit{P}\left(\mathit{x}=1\right)+\mathit{P}\left(\mathit{x}=2\right)$
$\Rightarrow {}^3{\mathit{C}}_0{\left(\frac{1}{36}\right)}^0{\left(\frac{35}{36}\right)}^3+{}^3{\mathit{C}}_1{\left(\frac{1}{36}\right)}^1\times {\left(\frac{35}{30}\right)}^2$ $+{}^3{\mathit{C}}_2{\left(\frac{1}{36}\right)}^2\left(\frac{35}{36}\right)$
$\Rightarrow 1-\frac{1}{{36}^3}$