Question

An ordinary cube has four blank faces, one face marked $2$ another marked $3$. The probability of obtaining a total of exactly $12$ in $5$ throws is $\frac{k}{6^5}$, then $k$ is

Answer 1

To obtain a total of $12$ in $5$ throws can be obtained in two ways
$\left(1\right)$ One blank and four $3^{\prime }s$ i.e., $(0,3,3,3,3)$
Number of ways ${}^5{C}_1=5$

$\left(2\right)$ Three $2^{\prime }s$ and two $3^{\prime }s$ i.e., $(2,2,2,3,3)$
Number of ways ${}^5{C}_2=10$

$\therefore P=\left(\frac{4}{6}\right)\times {\left(\frac{1}{6}\right)}^4\times {}^5{C}_1+{\left(\frac{1}{6}\right)}^5\times {}^5{C}_2=\frac{20}{6^5}+\frac{10}{6^5}=\frac{30}{6^5}$

$\Rightarrow k=30$