Question

Three numbers are chosen at random, one after another with replacement, from the set $S=\{1,2,3,...,100\}.$ Let $p_2$ be the probability that the minimum of chosen numbers is at most $40.$ Then the value of $\frac{125}{4}{p}_2$ is

Answer 1

$p_2=$probability that minimum of chosen numbers is at most $40$
$\Rightarrow p_2=1-$ probability that minimum of chosen numbers is at least $41$
$\Rightarrow p_2=1-{\left(\frac{60}{100}\right)}^3=\frac{98}{125}$
So, $\frac{125}{4}{p}_2=\frac{125}{4}\times \frac{98}{125}=\frac{98}{4}$
$\therefore \frac{125}{4}{p}_2=24.50$