A bag contains 5 white 7 red and 8 black balls. If four balls are drawn one by one with replacement what is the probability that (i) none is the white (ii) all are white (ii) at least one is white ?

Answer

`(i) .(81)/(256) (ii) .(1)/(256) (iii) (175)/(256)`

Explanation

P(white ) `=(5)/(20) =(1)/(4)` P(nonwhite) `=(3)/(4)`
`:. P =(1)/(4) ,q=(3)/(4)` and n=4
`P(X=r) =.^(n)C_(r ) .p^(r ).q^((n-r)) =.^(4)C_(r ).((1)/(4))^(r ) .((3)/(4))^((4-r))`
`(i) P(X =0) =.^(4)C_(0) .((3)/(4))^(4)`
`(ii) P(X=4) =.^(4)C_(4).((1)/(4))^(4)`
`(iii) P(Xge1)=1-P(Xlt1) =1- P(X =0)`