I ran across the following problem while practicing for the GRE math subject test:
Suppose $X$ is a discrete random variable on the set of positive integers such that for each positive integer $n$, the probability that $X = n $ is $\frac{1}{2^n}$. If $Y$ is a random variable with the same probability distribution and $X$ and $Y$ are independent, what is the probability that the value of at least one of the variables $X$ and $Y$ is greater than $3$?
In trying to solve this, I tried the following:
We want $P(X>3 \wedge Y>3) = 2P(X>3) = 2\sum_{i=4}^{\infty}\frac{1}{2^i} = \frac{1}{4}$. However, the correct answer is $\frac{15}{64}$ and I understand the reasoning for this answer, but why is mine incorrect? Where did I go wrong? Is it because I forgot to take into account the possibility that $X$ and $Y$ may not be disjoint (i.e. I am double counting), which seems possible since $\frac{15}{64}<\frac{1}{4}$.
You want to find $P(X > 3\: \text{or}\: Y > 3)$, which is the same as $1 - P(X \le 3\: \text{and}\: Y \le 3)$. Since $X$ and $Y$ have the same distribution and are independent, $P(X \le 3\: \text{and}\: Y \le 3) = P(X \le 3)P(Y\le 3) = P(X \le 3)^2$. So you compute
$$1 - P(X \le 3)^2 = 1 - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8}\right)^2 = 1 - \left(\frac{7}{8}\right)^2 = \frac{15}{64}.$$
The problem in your solution is the first step: $P(X > 3\: \text{or}\: Y > 3) \neq 2P(X > 3)$. Recall the rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.