Question

Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3,\dots ..,100\}$. 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

Answer: 76.25

$P_1=1$ - (Probability that $3$ chosen numbers are less than $81$ )
$=1-{\left(\frac{80}{100}\right)}^3=1-\frac{64}{125}=\frac{61}{125}$
So, $\frac{625}{4}$
$P_1=\frac{625}{4}\times \frac{61}{125}=76.25$