The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05 . Find the probability that out of 5 such bulbs
(i) none will fuse after 6 months of use
(ii) at least one will fuse after 6 months of use
(iii) not more than one will fuse after 6 months of use.

Answer

`(i) ((19)/(20))^(5) (ii) 1-((19)/(20))^(5) .(iii) .((19)/(20))^(4) xx ((6)/(5))`

Explanation

`p=(1)/(20) ,q=(1-(1)/(20)) =(19)/(20) " and ' n=5`
`P(X=r) =.^(n)C_(r ).p^(r ) .q^((n-r)) =.^(5)C_( r) .((1)/(20))^(r ).((19)/(20))^((5-r))`
`(i) P(x=0) =.^(5)C_(0) .((19)/(20))^(5)`
`(ii) P(X ge 1) = 1 -P(X lt 1) =1- P(X=0)`
`=1-.^(5)C_(0) .((19)/(20))^(5)`
`(iii) P(X le 1) =P(X=0) + P(X=1)`
`=.^(5)C_(0) .((19)/(20))^(5) +.^(5)C_(1).((1)/(20)) ((19)/(20))^(4)`