In a box containing 60 bulbs 6 are defective . What is the probability that out of a sample of 5 bulbs (i) none is defective (ii) exactly 2 are defective ?

Answer

`(i) ((9)/(10))^(5) (ii) (729)/(10000)`

Explanation

`p=(6)/(60)=(1)/(10),q =(1-(1)/(10)) =(9)/(10) " and " n=5`
`P(X=r) =.^(n)C_(r ).p^(r ).q^((n-r)) =.^(5)C_(r ) .((1)/(10))^(r ).((9)/(10))^((5-r))`
(i) P(none is defective ) `=P(X=0) =.^(5)C_(0) .((9)/(10))^(5)`
(ii) P(exactly 2 are defective ) `P(X=2) =.^(5)C_(2).((1))/(10))^(2) .((9)/(10))^(3)`