The items produced by a company contain 10% defective items. Show that the proximity of getting 2 defective items in a sample of 8 items is `(28xx9^6)/(10^8)` .

Explanation

Solution:

Let X denote the number of defective items in the items produced by the company. Then, X follows binomial distribution with n=`8`.

`p=10 %=frac{1}{10} `

`q=1-p=frac{9}{10}`


Hence, the distribution is given by


`P(X=r)=^{8} C_{r}(frac{1}{10})^{r}(frac{9}{10})^{8-r}`

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