Question

A factory produces bulbs. The probability that any one bulb is defective is $\frac{1}{50}$ and they are packed in 10 boxes. From a single box, find the probability that

(i) None of the bulbs is defective.

(ii) Exactly two bulbs are defective.

Answer 1

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

Also, $n=10,p=\frac{1}{50}$ and $q=\frac{49}{50}$ and $P(X=r)={}^n{C}_r{p}^r{q}^{n-r}$

(i) None of the bulbs is defective i.e., r = 0

$\therefore p(X=r)=P_{\left(0\right)}={}^{10}{C}_0\left(\frac{1}{50}{)}^0\right(\frac{49}{50}{)}^{10-0}=(\frac{49}{50}{)}^{10}$

(ii) Exactly two bulbs are defective i.e., r = 2

$\therefore P(X=r)=P_{\left(2\right)}={}^{10}{C}_2\left(\frac{1}{50}{)}^2\right(\frac{49}{50}{)}^8$

$=\frac{10!}{8!2!}(\frac{1}{50}{)}^2.(\frac{49}{50}{)}^8=45\times (\frac{1}{50}{)}^{10}\times (49{)}^8$

$={}^{10}{C}_0\left(\frac{1}{50}{)}^0\right(\frac{49}{50}{)}^{10-0}+{}^{10}{C}_1\left(\frac{1}{50}{)}^1\right(\frac{49}{50}{)}^{10-1}$

$=(\frac{49}{50}{)}^{10}+\frac{10!}{1!9!}.\frac{1}{50}.(\frac{49}{50}{)}^9$

$=(\frac{49}{50}{)}^{10}+\frac{1}{5}.(\frac{49}{50}{)}^9=\left(\frac{49}{50}{)}^9\right(\frac{49}{50}+\frac{1}{5})$

$=\left(\frac{49}{50}{)}^9\right(\frac{59}{50})=\frac{59(49{)}^9}{(50{)}^{10}}$