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

N/a

Explanation

`p=5/100=1/20`
and `q=1-p=1-1/20=19/20`
Therefore, the distribution `X` is a binomial distribution with `n=4,p=1/20` and `q=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).(1/20)^(r)(19/20)^(10-r)`
(Using binomial distribution)
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)+10pq^(9)`
`=q^(9)(q+10p)`
`=(19/20)^(9)(19/20+10xx1/20)=29/20(19/20)^(9)`