Question

Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3,\cdots ,100\}.$ Let $p_1$ be the probability that the maximum of chosen numbers is at least $81.$ Then the value of $\frac{625}{4}{p}_1$ is

Answer 1

$p_1=$probability that maximum of chosen numbers is at least $81$
$\Rightarrow p_1=1-$ probability that maximum of chosen number is atmost $80$
$\Rightarrow p_1=1-{\left(\frac{80}{100}\right)}^3=\frac{61}{125}$
So, $\frac{625}{4}{p}_1=\frac{625}{4}\times \frac{61}{125}=\frac{305}{4}$
$\therefore \frac{625}{4}{p}_1=76.25$