Question

There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Answer 1

Let X denote the number of defective items is a simple of items drawn succesively, Since, the drawing is done with replacement, the trails are Bernoulli trails.

$\therefore$ p=P (success) = 5% = $\frac{5}{100}=\frac{1}{20}$ and q=1- p=1 - $\frac{1}{20}=\frac{19}{20}$

X has a binomial distrubtion with n=10 and p=$\frac{1}{20}and=\frac{19}{20}$

Therefore, P(X=r)=${}^n{C}_r{p}^r{q}^{n-r}$ , where r= 0,1,2,....,n

p(X=r)= ${}^{10}{C}_r{\left(\frac{1}{20}\right)}^r{\left(\frac{19}{20}\right)}^{10-r}$ (By Binomial distrubtion)

Required probability = P (not more than one defective item)

= P(0)+P(1)= $={}^{10}{C}_0{p}^0{q}^{10}+{}^{10}{C}_1{p}^1{q}^9=1q^{10}+10p{q}^9$

=$=q^9(q+10p)={\left(\frac{19}{20}\right)}^9(\frac{19}{20}+10\times \frac{1}{20})=\frac{29}{20}{\left(\frac{19}{20}\right)}^9$