A random variable X takes the values `0,1,2,3,...,` with prbability `PX(=x)=k(x+1)((1)/(5))^x`, where k is a constant, then `P(X=0)` is.

Options

A. `7/(25)`
B. `(18)/(25)`
C. `(13)/(25)`
D. `(16)/(25)`

Answer

D

Explanation

`P(X=0)=k,P(X=1)=2k(1/5)^1`
`P(X=2)=3k(1/5)^2...=1`
Since `P(X=0)+P(X=1)+P(X=2)+...=1`
`thereforek+2k(1/5)+3k(1/5)^2+...=1`
`and (k/5+2k(1/5)^2+...=1/5)/(k+k(1/5)+k(1/5)^2+...=4/5)`
`rArr" "k/(1-1/5)=4/5rArrk=(16)/(25)`
`therefore" "P(X=0)=(16)/(25)(0+1)(1/5)^0=(16)/(25)`