The mean and variance of a binomial distribution are 4 and 4/3 respectively, find `P(Xgeq1)dot`

Answer

`(728)/(729)`

Explanation

`(np=4 " and " npq =(4)/(3)) rArr q=(1)/(3)`
`:. , p =(1-q)= (1-(1)/(3)) =(2)/(3) " and " n=(4)/(p) =(4 xx (3)/(2))=6`
`:. P(X =r) =.^(n)C_( r).p^(r ) .q^((n-r))`
`=.^(6)C_( r).((2)/(3))^(r ).((1)/(3))^((6-r))`
`P(Xge1) =1- P(X=0)`
`=1-.^(6)C_(0).((2)/(3))^(0).((1)/(3))^(6)`
`=(1-(1)/(3^(6)))=(728)/(729)`