Question

A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that
(i) none of the bulbs is defective
(ii) exactly two bulbs are defective
(iii) more than 8 bulbs work properly [NCERT EXEMPLAR]

Answer 1

Let getting a defective bulb from a single box is a success.

We have,

$$\begin{gathered} p=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{defective}\mathrm{bulb}=\frac{1}{50} \\ \mathrm{Also},q=1-p=1-\frac{1}{50}=\frac{49}{50} \\ \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{success}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}10\mathrm{trials}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=10\mathrm{and}p=\frac{1}{50} \\ \therefore \mathrm{P}\left(X=r\right)={}^{10}\mathrm{C}_r{p}^r{q}^{\left(10-r\right)}={}^{10}\mathrm{C}_r{\left(\frac{1}{50}\right)}^r{\left(\frac{49}{50}\right)}^{\left(10-r\right)}=\frac{{}^{10}\mathrm{C}_r{49}^{\left(10-r\right)}}{{50}^{10}},\mathrm{where}r=0,1,2,3,...,10 \\ \mathrm{Now}, \\ \left(\mathrm{i}\right)\mathrm{Required}\mathrm{probability}=\mathrm{P}\left(\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{bulb}\mathrm{is}\mathrm{defective}\right) \\ =\mathrm{P}\left(X=0\right) \\ =\frac{{}^{10}\mathrm{C}_0{49}^{10}}{{50}^{10}} \\ =\frac{{49}^{10}}{{50}^{10}} \\ ={\left(\frac{49}{50}\right)}^{10} \\ \left(\mathrm{ii}\right)\mathrm{Required}\mathrm{probability}=\mathrm{P}\left(\mathrm{exactly}\mathrm{two}\mathrm{bulbs}\mathrm{are}\mathrm{defective}\right) \\ =\mathrm{P}\left(X=2\right) \\ =\frac{{}^{10}\mathrm{C}_2{49}^8}{{50}^{10}} \\ =\frac{45\times {49}^8}{{50}^{10}} \\ \left(\mathrm{iii}\right)\mathrm{Required}\mathrm{probability}=\mathrm{P}\left(\mathrm{more}\mathrm{than}8\mathrm{bulbs}\mathrm{work}\mathrm{properly}\right) \\ =\mathrm{P}\left(\mathrm{atmost}\mathrm{one}\mathrm{bulb}\mathrm{is}\mathrm{defective}\right) \\ =\mathrm{P}\left(X\le 0\right) \\ =\mathrm{P}\left(X=0\right)+\mathrm{P}\left(X=1\right) \\ =\frac{{}^{10}\mathrm{C}_0{49}^{10}}{{50}^{10}}+\frac{{}^{10}\mathrm{C}_1{49}^9}{{50}^{10}} \\ =\frac{{49}^{10}}{{50}^{10}}+\frac{10\times {49}^9}{{50}^{10}} \\ =\frac{{49}^9}{{50}^{10}}\left(49+10\right) \\ =\frac{59\left({49}^9\right)}{\left({50}^{10}\right)} \end{gathered}$$