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)` ,brgt `P(X=2)=3k((1)/(5))^(2)` , . .
since, `P(X=0)+P(X=1)+P(X=2)+ . . =1`
`thereforek+2k((1)/(5))+3k((1)/(5))^(2)+ . . .=1` . . (i)
On multiplying both sides by `(1)/(5')`
we get `(k)/(5)+2k((1)/(5))^(2)+. . .=(1)/(5)` . . . (ii)
Ob subtracting Eq. ((i) from (ii), we get
`k+k((1)/(5))+k((1)/(5))^(2)+ . . .=(4)/(5)`
`implies(k)/(1-(1)/(5))=(4)/(5)impliesk=(16)/(25)`
`thereforeP(X=0)=(16)/(25)(0+1)((1)/(5))^(0)=(16)/(25)`.