Question

In a large bulk of items, 5 per cent of the items are defective. 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 in a sample of 10 items.

X follows a binomial distribution with n =10; $p\mathit{}=\text{p}\mathrm{robability}\text{of}\mathrm{defective}\mathrm{item}s=5\%=0.05;q\mathit{}=1-p=0.95$

$$\begin{gathered} P(X=\mathit{\text{r}})={}^{10}C_r(0.05{)}^r(0.95{)}^{\mathit{10}\mathit{-}r} \\ \mathrm{Probability}(\mathrm{sample}\mathrm{of}10\mathrm{items}\mathrm{will}\mathrm{include}\mathrm{not}\mathrm{more}\mathrm{than}\mathrm{one}\mathrm{defective}\mathrm{item})=\mathit{}P(X\le 1) \\ =P(X=0)+P(X=1) \\ ={}^{10}C_0(0.05{)}^0(0.95{)}^{10-0}+{}^{10}\mathrm{C}_1(0.05{)}^1(0.95{)}^{10-1} \\ =(0.95{)}^9(0.95+0.5) \\ =1.45(0.95{)}^9 \end{gathered}$$

$={\left(\frac{19}{20}\right)}^9\frac{29}{20};(\because 1.45=\frac{29}{20}\mathrm{and}0.95=\frac{19}{20})$