Question

If the probability that the product of the outcomes of three rolls of a fair dice is a prime number is $p,$ then the value of $\frac{1}{\left(4p\right)}$ is

Answer 1

Total number of cases $n\left(S\right)=6^3=216$
Product is prime, only when two outcomes are $1$ and third is prime i.e. $2,3,5$
If it is $2,1,1,$ then the number of cases is $3$
Similarly, $3$ cases for $3,1,1$ and $5,1,1$ each
Hence, favorable cases $=9$
Therefore, required probability $\left(p\right)=\frac{9}{216}=\frac{1}{24}$
$\Rightarrow \frac{1}{4p}=6$