Question

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 _____.

Answer 1

$P_{1}=1$ - (Probability that $3$ chosen numbers are less than $81$ )
$=1-\left(\frac{80}{100}\right)^{3}=1-\frac{64}{125}=\frac{61}{125}$
So, $\frac{625}{4}$
$P_{1}=\frac{625}{4} \times \frac{61}{125}=76.25$