How to find probability in when using an "or" statement Can somebody please help me solve/understand the following problem?:
Consider two random variables $X$ and $Y$ with events $S_{X}=S_{Y}=\{1,2,3,4\}$ and probabilities:
$P(X=j)=\frac{1}{4} \quad$ for all $j \in \{1,2,3,4\}, \quad$ and $\quad P(Y=j)=\left\{\begin{array}{ll}\frac{1}{8}, & \text { if } j=1 \\ \frac{1}{4}, & \text { if } j=2 \\ \frac{3}{8}, & \text { if } j=3 \\ \frac{1}{4}, & \text { if } j=4\end{array}\right.$
Assume, that $X$ and $Y$ are independent.

*

*Find the probablity $P(X \leq 2$ or $Y \leq 2)$.


First I tried to solve it like this:
$P(x \leq 2$ or $y \leq 2)=P(x \leq 2)+P(y \leq 2)=\frac{7}{8}$
But I realised that this method would give a probability higher than 1 if I considered $x\leq4$ or $y\leq 4$. So how do I solve this correctly, and why is this method wrong?
One thing, that I'm not sure of, is whether I should still count the possibilities when X and Y are $\leq2$ at the same time? Or if it can only be one of them at a time, for the statement to be true?
Secondly, I found this theorem in my book, that might be helpful (see below). Specifically, I think a) could be used? But I just can't seem to "translate" the theorem to this problem. The first problem is, that my theorem only considers one random variable, but my case has two. Secondly, I'm not certain how to use the union and intersection of sets in this problem. But if it's possible to use this theorem, I would really appreciate you helping me understanding how.
Thanks in advance

Theorem: Let $X$ be a random variable with the possible events $S_{X},$ and let $A$ and $B$ be subsets of $S_{X}$. Then the following rules can be used:
a. $P(X \in A \cup B)=P(X \in A)+P(X \in B)-P(X \in A \cap B)$
b. $P(X \in A \cup B) \leq P(X \in A)+P(X \in B)$.
c. If $A$ and $B$ are disjoint, then
$
P(X \in A \cup B)=P(X \in A)+P(X \in B)
$
d. $P(X \notin A)=P\left(X \in S_{X} \backslash A\right)=1-P(X \in A)$
e. If $A_{1}, A_{2} \ldots, A_{n}$ are disjoint subsets of $S_{X},$ then
$  
P\left(X \in A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)=P\left(X \in A_{1}\right)+P\left(X \in A_{2}\right)+\cdots+P\left(X \in A_{n}\right)
$
 A: For your answer to be correct, you need to make sure you are adding probabilities of disjoint events. So you should look at it as only $\small X \leq 2$ occurs $+$ only $\small Y \leq 2$ occurs $+$ both occur.
So in your working, it should have been $\frac{1}{2} \times (1 - \frac{3}{8}) + \frac{3}{8} \times (1 - \frac{1}{2}) + \frac{3}{8} \times \frac{1}{2} = \frac{11}{16}$
If you want to apply the formula you showed later on,
$\small P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$
Please note $\small P(X \cap Y) = P(X) \, P(Y) \,$  for $X, Y \,$ are independent.
$ \small P(X \cup Y) = \frac{1}{2} + \frac{3}{8} - \frac{1}{2} \times \frac{3}{8}$
Or you could simply use the complement events to find the desired probability as,
$ \, \small P(X \cup Y) = 1 - P(X^c \cap Y^c) = 1 - (1 - \frac{1}{2}) \times (1 - \frac{3}{8})$
A: You cannot use the theorem as it is, as for using the additivity you tried to use you have to know that $A$ and $B$ are disjoint, which you do not know.
You can use the fact that for independent variables $\mathbb{P}(A\cap B) = \mathbb{P}(A) \mathbb{P}(B)$. With $A^c$ denoting the complement of $A$ you can write $\mathbb{P}(A \cup B) = \mathbb{P}(((A\cup B)^c)^c)= \mathbb{P}((A^c\cap B^c)^c) = 1- \mathbb{P}(A^c\cap B^c)$.
If you just plug in $A= \{X\leq 2\}$ and $B= \{Y \leq 2\}$ you can calculate the answer.
