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  more than 8 bulbs work properly                                                                                                                            

 

 

Answer 1

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

$$p=\text{ probability of getting a defective bulb}=\frac{1}{50}$$
$$\text{ Also,}q=1-p=1-\frac{1}{50}=\frac{49}{50}$$
$$\text{ Let X denote the number of success in a sample of 10 trials . Then, }$$
$$\text{ X follows binomial distribution with parameters n = 10 and p }=\frac{1}{50}$$
$$\therefore P(X=r){=}^{10}{C}_r{p}^r{q}^{(10-r)}{=}^{10}{C}_r{\left(\frac{1}{50}\right)}^r{\left(\frac{49}{50}\right)}^{(10-r)}=\frac{{}^{10}{C}_r{49}^{(10-r)}}{{50}^{10}},\text{ where }r=0,1,2,3,...,10$$
$$\text{ Now},$$

$$\text{ Required probability }=P\left(\text{ more than 8 bulbs work properly }\right)$$
$$=P\left(\text{ atmost one bulb is defective}\right)$$
$$=P(X\le 0)$$
$$=P(X=0)+P(X=1)$$
$$=\frac{{}^{10}{C}_0{49}^{10}}{{50}^{10}}+\frac{{}^{10}{C}_1{49}^9}{{50}^{10}}$$
$$=\frac{{49}^{10}}{{50}^{10}}+\frac{10\times {49}^9}{{50}^{10}}$$
$$=\frac{{49}^9}{{50}^{10}}(49+10)$$
$$=\frac{59\left({49}^9\right)}{\left({50}^{10}\right)}$$