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  • A factory produces bulbs. The probability that any one bulb is defective is 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 bu

A factory produces bulbs. The probability that any one bulb is defective is 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

Answers (1)

Let X be the random variable which denotes that the bulb is defective.

And n=10, p=\frac{1}{50}$ and $P(X=r)=n_{c_{r}}(p)^{r} q^{n-r}$
(j) None of the bulbs is defective i.e., r=0 
\therefore P(X=r)=P_{0}=10_{c_{0}}\left(\frac{1}{50}\right)^{0}\left(\frac{49}{50}\right)^{10-0}=\left(\frac{49}{50}\right)^{10}$
(ii)Exactly two bulbs are defective i.e., r=2 
\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{P}_{2}=10 \mathrm{c}_{2}\left(\frac{1}{50}\right)^{2}\left(\frac{49}{50}\right)^{10-2}

\begin{aligned} &=\frac{10 !}{8 ! 2 !}\left(\frac{1}{50}\right)^{2} \cdot\left(\frac{49}{50}\right)^{8}\\ &=45 \times\left(\frac{1}{50}\right)^{10} \times(49)^{8}\\ &\text { (iii)More than } 8 \text { bulbs work properly i.e., there are less than } 2 \text { bulbs that are defective. }\\ &\text { Therefore, } r<2 \Rightarrow r=0,1\\ &\therefore P(X=r)=P(r<2)=P(0)+P(1)\\ &=10_{c_{0}}\left(\frac{1}{50}\right)^{0}\left(\frac{49}{50}\right)^{10-0}+10_{c_{1}}\left(\frac{1}{50}\right)^{1}\left(\frac{49}{50}\right)^{10-1}\\&=\left(\frac{49}{50}\right)^{10}+\frac{1}{5} \times\left(\frac{49}{50}\right)^{9}\\ &=\left(\frac{49}{50}+\frac{10}{50}\right)\left(\frac{49}{50}\right)^{9}\\ &=\frac{59(49)^{9}}{(50)^{10}} \end{aligned}

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