A random variable X takes values 0, 1, 2, 3,… with probability proportional to `(x+1)((1)/(5))^x`.
`P(X=0)` equals

Options

A. `(7)/(25)`
B. `(16)/(25)`
C. `(18)/(25)`
D. `(19)/(25)`

Answer

B

Explanation

The Probability distribution of X is
` P(X=x)=k(x+1)((1)/(5))^x,x=0,1,2,3,....`
`therefore sum _(x=0)^(oo)P(X=x)=1`
`rArr k sum _(x=0)^(oo)(x+1)((1)/(5))^x=1`
`rArr k{1+2xx((1)/(5))+3xx((1)/(5))^2+4xx((1)/(5))^3+.....}=1`
`rArr k{(1)/(1-(1)/(5))+(1xx(1)/(5))/((1-(1)/(5))^2)}=1[because a +(a+d)r+(a+2d)r^2+.......=(a)/(1-r)+(dr)/(1-r)^2]`
`rArr kxx(25)/(16)=1rArrk=(16)/(25)` .