($1$) 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 $76.25$
($2$) $p _2=\text { probability that minimum of chosen numbers is at most } 40$
$=1-\text { probability that minimum of chosen numbers is at least } 41$
$=1-\left(\frac{600}{100}\right)^3$
$=1-\frac{27}{125}=\frac{98}{125}$
$\therefore \frac{125}{4} p _2=\frac{125}{4} \times \frac{98}{125}=24.50$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
જો $F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{\left(e^{\beta}-4\right)^{2}}$ તો $\alpha+\beta$ ની કિમંત મેળવો.