There are 6% defective items in a large bulk of item. Find the probability that  sample of 8 items will include not more than one defective items.

Explanation

Using Bernoulli's Trial `P(`Success`=x) =n_{C_{x}} cdot p^{x} cdot q^{(n-x)} x=0,1,2, ldots ldots ldots n and q=(1-p), n=8`

The probability of success, i.e. the bulb is defective `=p=frac{6}{100}=frac{6}{100} q=1-frac{6}{100}=frac{94}{100}`

probability of that there is not more than one defective piece` =
P(0` defective items `)+P(1` defective item )`=

^{8} {C}_{0} cdot(frac{6}{100})^{0}(frac{94}{100})^{8}+{ }^{8} {C}_{1} cdot(frac{6}{100})^{1}(frac{94}{100})^{7} `

`Rightarrow((frac{47}{50})^{7} times(frac{71}{50}))`