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 will fuse after 150 days of use
(ii) not more than one will fuse after 150 days of use
(iii) more than one will fuse after 150 days of use
(iv) at least one will fuse after 150 days of use

Answer 1

Let X be the number of bulbs that fuse after 150 days.

X follows a binomial distribution with n = 5, $p\mathit{}=0.05\mathrm{and}q=0.95$
$$\begin{gathered} \text{Or}p=\frac{1}{20}\mathrm{and}q=\frac{19}{20} \\ P(X=r)={}^{5}C_r{\left(\frac{1}{20}\right)}^r{\left(\frac{19}{20}\right)}^{\mathit{5}\mathit{-}r} \\ \left(\mathrm{i}\right)\mathrm{Probability}(\mathrm{none}\mathrm{will}\mathrm{fuse}\mathrm{after}150\mathrm{days}\mathrm{of}\mathrm{use})=P(X=0) \\ ={}^{5}C_0{\left(\frac{1}{20}\right)}^0{\left(\frac{19}{20}\right)}^{5-0} \\ ={\left(\frac{19}{20}\right)}^5 \\ \left(\mathrm{ii}\right)\mathrm{Prob}\text{ability}(\mathrm{not}\mathrm{more}\mathrm{than}1\mathrm{will}\mathrm{fuse}\mathrm{after}150\mathrm{days}\mathrm{of}\mathrm{us}e)=\mathit{}P(X\le 1) \\ =\mathit{}P(X=0)+P(\mathrm{X}=1) \\ ={\left(\frac{19}{20}\right)}^5+5{\mathrm{C}}_1{\left(\frac{1}{20}\right)}^1{\left(\frac{19}{20}\right)}^{5-1} \\ ={\left(\frac{19}{20}\right)}^4\left\{\frac{19}{20}+\frac{5}{20}\right\} \\ =\frac{6}{5}{\left(\frac{19}{20}\right)}^4 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Probability}\left(\mathrm{more}\mathrm{than}\mathrm{one}\text{will}\mathrm{fuse}\mathrm{after}150\mathrm{days}\mathrm{of}\mathrm{use}\right)=P(X>1) \\ =1-P(X\le 1) \\ =1-\frac{6}{5}{\left(\frac{19}{20}\right)}^4\left\{\because P(X\le 1)=\frac{6}{5}{\left(\frac{19}{20}\right)}^4\right\} \\ \left(\mathrm{iv}\right)\mathrm{Probability}\left(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{will}\mathrm{fuse}\mathrm{after}150\mathrm{days}\mathrm{of}\mathrm{use}\right)=P(X\ge 1) \\ =1-P(X=0) \\ =1-{\left(\frac{19}{20}\right)}^5\left\{\because P(X=0={\left(\frac{19}{20}\right)}^5\right\} \end{gathered}$$