Question

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one
will fuse after 150 days of use.

Answer 1

Let $X$ represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.
It is given that, $p=0.05$
$\therefore q=1-p=1-0.05=0.95$
$X$ has a binomial distribution with $n=5$ and $p=0.05$
$\therefore P(X=x){=}^n{C}_x{q}^{n-x}{p}^x$, where $x=1,2,...n$
${=}^5{C}_x{\left(0.95\right)}^{5-x}\cdot {\left(0.05\right)}^x$

(i) $P\left(none\right)=P(X=0)$
${=}^5{C}_0{\left(0.95\right)}^5\cdot {\left(0.05\right)}^0$
$=1\times {\left(0.95\right)}^5$
$={\left(0.95\right)}^5$

(ii) $P\left(notmorethanone\right)=P(X\le 1)$
$=P(X=0)+P(X=1)$
${=}^5{C}_0{\left(0.95\right)}^5\times {\left(0.05\right)}^0{+}^5{C}_1{\left(0.95\right)}^4\times {\left(0.05\right)}^1$
$=1\times {\left(0.95\right)}^5+5\times {\left(0.95\right)}^4\times \left(0.05\right)$
$={\left(0.95\right)}^5+\left(0.25\right){\left(0.95\right)}^4$
$={\left(0.95\right)}^4[0.95+0.25]$
$={\left(0.95\right)}^4\times 1.2$

(iii) $P\left(morethan1\right)=P(X>1)$
$=1-P(X\le 1)$
$=1-P\left(notmorethan1\right)$
$=1-{\left(0.95\right)}^4\times 1.2$

(iv) $P\left(atleastone\right)=P(X\ge 1)$
$=1-P(X<1)$
$=1-P(X=0)$
$=1{-}^5{C}_0{\left(0.95\right)}^5\times {\left(0.05\right)}^0$
$=1-1\times {\left(0.95\right)}^5$
$=1-{\left(0.95\right)}^5$