The question is asking for values for $k$ making this a valid probability mass function.
where $$P(x) = k(1-r^2)^x$$
for $x = 0,1,2...$ and $P(x)= 0$ otherwise
$r$ is elected from interval $(0,1)$
I'm thinking this is a discrete probability distribution that needs to sum to 1, but the text is very light on and I'm at a bit of a loss how to solve this.
Any help appreciated
Summing over $x$, $$1=\sum_{x=0}^\infty P(x) = k\sum_{x=0}^\infty (1-r^2)^x = \frac{k}{1-(1-r^2)}$$ by the geometric series formula. So $k=r^2$.
If you recognize it as a geometric pmf
$$P(X=x) = p(1-p)^x, \: x=0,1,2,...$$
with $p=r^2$ then it's immediately obvious that $k=p$.
Of course if you don't recognize the pmf then it's at least useful to recognize that it's a geometric series.
