($1$) The value of $\frac{625}{4} p _1$ is
($2$) The value of $\frac{125}{4} p _2$ is
Give the answer or queution ($1$) and ($2$)
$p _1=1-\text { probability that maximum of chosen number is at most } 80$
$p _1=1-\frac{80 \times 80 \times 80}{100 \times 100 \times 100}=1-\frac{64}{125}$
$p _1=\frac{61}{125}$
$\frac{625 p _1}{4}=\frac{625}{4} \times \frac{61}{125}=\frac{305}{4}=76.25$
the value of $\frac{625 p _1}{}$ is $76.25$
($2$) $p _2=\text { probability that minimum of chosen numbers is at most } 40$
$=1-\text { probability that minimum of chosen numbers is at least } 41$
$=1-\left(\frac{600}{100}\right)^3$
$=1-\frac{27}{125}=\frac{98}{125}$
$\therefore \frac{125}{4} p _2=\frac{125}{4} \times \frac{98}{125}=24.50$
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