Question

Counters numbered $1,2,3$ are placed in a bag and one is drawn at random and replaced. The operation is being repeated three times. If the probability of obtaining a total $6,$ is $p.$ Then the value of $\left[\frac{1}{p}\right]$ is
(where $[\cdot ]$ denotes greatest integer function)

Answer 1

To get a sum of $6,$ the possible combination of selection is $(2,2,2)$ or $(1,2,3)$
So Required probability
$=P(2,2,2)+3!\times P(1,2,3)=\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}+6(\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3})=\frac{7}{27}$
$\Rightarrow p=\frac{7}{27}\therefore \left[\frac{1}{p}\right]=3$