Question

The probability that a bulb produced by a factory will fuse in $100$ days of use is $0.05$. Find the probability that out of $5$ such bulbs, after $100$ days of use, more than one fuses.

Answer 1

$p=0.05=\frac{1}{20},q=\frac{19}{20}$
P(Not more than one fuses)
$={\sum }_{r=0}^1{}^n{C}_r{p}^r{q}^{n-r}$
${=}^5{C}_0{\left(\frac{1}{20}\right)}^0{\left(\frac{19}{20}\right)}^5{+}^5{C}_1\left(\frac{1}{20}\right){\left(\frac{19}{20}\right)}^4$
$={\left(\frac{19}{20}\right)}^4(\frac{19}{20}+\frac{5}{20})={\left(\frac{19}{20}\right)}^4\left(\frac{24}{20}\right)=\frac{6}{5}{\left(\frac{19}{20}\right)}^4$

P(More than one fuses) $=1-$P(Not more than one fuse)
$=1-\frac{6}{5}{\left(\frac{19}{20}\right)}^4$