I think I have the hang of this now, but I have a minor question:

I am to find the constant $\displaystyle c$ which would make $\displaystyle f(x)$ satisfy all the conditions of being a p.m.f.

$\displaystyle f(x)=c(1/4)^x;~~~x=1,2,3,...$

I assumed that this would continue on to infinity [for there was another similar to this, where it explicitly said that $\displaystyle x=1,2,3...n$]

So, I figured that $\displaystyle f(x)>0$;

I also noticed that $\displaystyle \sum_{x\in S}f(x)=1\implies \sum_{x=1}^{\infty}c(1/4)^x=1$. The series is geometric, where $\displaystyle a=c$, and $\displaystyle r=\frac{1}{4}$. Thus, it converges to $\displaystyle \frac{c}{1-\frac{1}{4}}=\frac{c}{\frac{3}{4}}=\frac{4}{3}c$

So for $\displaystyle \frac{4}{3}c=1,~c=\frac{3}{4}$

However, my text tells me that the answer is 3.

Maybe I'm overlooking one simple thing...wait!

How do I apply the last property of a p.m.f:

$\displaystyle P(X\in A)=\sum_{x\in A}f(x)~~~\text{where}~A\subset S$

to this problem?

I'd appreciate any input!

--Chris