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.

Answer

`((46)/(50))^(7) xx ((71)/(50))`

Explanation

`p=(6)/(100) =(3)/(50) ,q=(1-(3)/(50)) =(47)/(50)" and " n=8`
` P(X=r) =.^(n)C_(r ) .p^(r ).q^((n-r)) =.^(8)C_(r ) .((3)/(50))^( r) .((47)/(50))^((8-r))`
Required probability =P( 0 defective or 1 defective )
`=P(X=0) + P(X=1)`
` =.^(8)C_(0).((47)/(50))^(8)+.^(8)C_(1).((3)/(50))((47)/(50))^(7)`