Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3, \ldots, 100\}$. Let $p_{1}$ be the probability that the maximum of chosen numbers is at least 81 and $p_{2}$ be the probability that the minimum of chosen numbers is at most 40 .
The value of $\frac{625}{4}{p}_1$ is

Maximum of the chosen numbers is at least $81$

It means we have to choose at least one number from $81$ to $100$

Total number of possible selections $=100\times 100\times 100={100}^3$

Favourable cases $=$ Total $-$ unfavourable cases

Unfavourable cases are those in which we have selected all the three numbers form $1$ to $80$ $=80\times 80\times 80={80}^3$

Total number of favourable cases $={100}^3-{80}^3$

So, $p_1=\frac{{100}^3-{80}^3}{{100}^3}$

$=\frac{{20}^3\left(5^3-4^3\right)}{{20}^3\times 5^3}=\frac{125-64}{125}=\frac{61}{125}$

Hence, $\frac{625}{4}{p}_1=\frac{625}{4}\times \frac{61}{125}=76.25$