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 $\frac{125}{4}{p}_2$ is ______.

Answer 1

Answer: 24.5

$P_2=1$-(Probability that 3 chosen numbers are greater than 40 )
$=1-{\left(\frac{60}{100}\right)}^3=1-{\left(\frac{3}{5}\right)}^3=\frac{98}{125}$
So, $\frac{125}{4}$
$P_2=\frac{125}{4}\times \frac{98}{125}=24.5$