Again from Princeton Review's fourth edition, chapter 7 review:

#38 Let X be a random variable on $\displaystyle \mathbb{Z}^+$ (positive integers) whose distribution function is $\displaystyle F_x (t) = \frac 1 {3^t}$. Suppose that Y is another random variable whose distribution function is $\displaystyle F_Y (t)= \frac 1 {4^t}$. What is the probability that at least one of the variables X and Y is greater than 2?

(a) 5/6

(b) 64/81

(c) 1/2

(d) 17/81

(e) 1/6

First off, the question seems poorly worded since I am not sure if Y is also a random variable on the positive integers. Furthermore, shouldn't their distribution functions be increasing? The answer in the book is (e). The explanation of the answer seems to treat the functions $\displaystyle F_X$ and $\displaystyle F_Y$ as though they are describing the probability that $\displaystyle X\ge t$ and $\displaystyle Y \ge t$ instead of, what I thought probability distribution functions described, $\displaystyle P(X\le t)$ and $\displaystyle P(Y \le t)$. Here is the explanation:

The problem asks for the probability that at least one of the variables is greater than 2. That's the same as 1 minus the probability that both variables are less than or equal to 2. Since X and Y are independent, we can multiply the complements of their individual probabilities. So

$\displaystyle \begin{align*}P(\text{at least one} >2) &= 1-P(X\le 2)P(Y\le 2)\\ &= 1-\left(\frac {3^2-1}{3^2} \right )\left(\frac{4^2-1}{4^2} \right )\\ &= 1-\frac 5 6\\ &= \frac 1 6\end{align*}$

I understand the logic of the first part, but why is it that they seem to say that $\displaystyle P(X \le 2)=1-\frac 1 {3^2}$? It seems to me that it would just be $\displaystyle \frac 1 {3^2}$, but then again those probability distribution functions don't make sense since they are decreasing...