If X follows a binomial distribution with mean 3 and variance `(3//2)` find
`(i) P(X ge 1) (ii) P(X le 5)`

Explanation

We know that mean =mp and variance = npq.
`:. , np=3 " and " npq =(3)/(2) rArr 3q = (3)/(2) rArr q= (1)/(2)`
`:. ,p=(1-q)=(1-(1)/(2))=(1)/(2)`
Now `np =3 " and " p=(1)/(2) rArr nxx (1)/(2) =3 rArr n=6`
So, the binomial distribuation is given by
`P(X=r) =.^(n)C_(r ).p^(r ) .q^((n-r)) =.^(6)C_( r) .((1)/(2))^(r ) .((1)/(2))^((6-r)) =.^(6)C_(r ).((1)/(2))^(6)`
`(i) P(X ge 1) =1- P(X=0)`
`=1-.^(6)C_(0) .((1)/(2))^(6) =(1-(1)/(64)) =(63)/(64)`
`(ii) P(X le 5) =1 - P(X=6)`
`=1-.^(6)C_(6) .((1)/(2))^(6) =(1-(1)/(64)) =(63)/(64)`