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,

$i=0\implies P(X=1)=\dfrac15P(X=0)=\dfrac p5$

$i=1\implies P(X=2)=\dfrac15P(X=1)=\dfrac p{5^2}$

and so on.

Then

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