Calculate Scaled Probability of Maximum Selected Number

To find the value of $\frac{625}{4} p_1$, we first analyze the given problem setup:

Let the set be $S = \{1, 2, 3, \dots, 100\}$, which contains $100$ elements. Three numbers $X_1, X_2, X_3$ are chosen at random, sequentially with replacement, from $S$.

We are given that $p_1$ is the probability that the maximum among the three chosen numbers is at least $81$, i.e., $p_1 = P(\max(X_1, X_2, X_3) \ge 81)$

Using the principle of complementary probability, we can express $p_1$ as: $p_1 = 1 - P(\max(X_1, X_2, X_3) \le 80)$

For the maximum of the three numbers to be less than or equal to $80$, each individual number selected must be less than or equal to $80$. That is, $P(\max(X_1, X_2, X_3) \le 80) = P(X_1 \le 80 \text{ and } X_2 \le 80 \text{ and } X_3 \le 80)$

Since the numbers are drawn independently with replacement: $P(X_i \le 80) = \frac{80}{100} = \frac{4}{5} \quad \text{for } i \in \{1, 2, 3\}$

Thus, the probability that all three numbers are at most $80$ is: $P(\max(X_1, X_2, X_3) \le 80) = \left(\frac{4}{5}\right)^3 = \frac{64}{125}$

Now, calculating $p_1$: $p_1 = 1 - \frac{64}{125} = \frac{61}{125}$

Finally, we calculate the required expression $\frac{625}{4} p_1$: $\frac{625}{4} p_1 = \frac{625}{4} \times \frac{61}{125} = \frac{5 \times 61}{4} = \frac{305}{4} = 76.25$

Thus, the numerical value of $\frac{625}{4} p_1$ is $76.25$.