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, at least one fuses.

Answer 1

$p=0.005=\frac{1}{20},q=\frac{19}{20}$

P(None fuse) $={}^5{C}_0{p}^0{q}^{5-0}={\left(\frac{1}{20}\right)}^0{\left(\frac{19}{20}\right)}^5$

P(At least one fuses) $=1-$ P(None fuses)

$=1-{\left(\frac{19}{20}\right)}^5$.