Question

suppose x is a non-negative integer valued random variable which follows a probability distribution such that p x=0 = 06 and for each i>=1, P(X=i+1) = 1/5 P(x=i). Calculate P(X<=2)

Answer 1

Not sure if the given probability that
`X=0` is 0.06 or 0.6, or something else altogether. I'll just refer to it by the number
`p`.

`\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge1\end{cases}`

As is, the probability that
`X=1` is indeterminate, so I think you intended to write

`\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge\boxed{0}\end{cases}`

Then by this definition,

`i=0\implies P(X=1)=\frac15P(X=0)=\frac p5`

`i=1\implies P(X=2)=\frac15P(X=1)=\frac p{5^2}`

and so on.

Then

`\boxed{P(X\le2)=P(X=0)+P(X=1)+P(X=2)=p+\frac p5+\frac p{5^2}=(31p)/(25)}`