The bulbs produced in a factory are supposed to contain 5% defective bulbs . What is the probability that a sample of 10 bulbs will contain not more than 2 defective bulbs?

Explanation

P(getting a defective bulb) `=(5)/(100) =(1)/(20)` and
P(getting a nondefective bulb) `=(1-(1)/(20)) =(19)/(20)`
Then `p =(1)/(20) " and " q=(19)/(20)`
Let X denote the number of defective bulbs
`P(X=r)=.^(n)C_(r ).p^(r).q^((n-r)) =.^(10)C_(r ).((1)/(20))^( r) .((19)/(20))^((10-r))`
P(getting not more than 2 defective bulbs)
`=P(X=0 " or " X=1 " or " X=2)`
`=P(X =0) +P(X=1)+P(X=2)`
`=.^(10)C_(0).((1)/(20))^(0).((19)/(20))^(10) +.^(10)C_(1).((1)/(20))^(1).((19)/(20))^(9)+.^(10)C_(2).((1)/(20))^(2).((19)/(20))^(8)`
`=((19)/(20))^(10)=(1)/(2).((19)/(20))^(9)+(9)/(80).((19)/(20))^(8)=((19)/(20))^(8) .((149)/(100))`
Let `A =((19)/(20))^(8) .((149)/(100)) .` Then
log A =8(log 19-log 20) +log 149 -log 100
`=8(1.2788 -1.3010 )+ 2.1732 -2 `
`=-0.0044 =1.9956`
`:. A =` antilog `(bar(1).9956)=0.99.`
Hence the required probability `=(99)/(100)`