have I thought correctly, $p(x\leq3\cup x>7) $

Question

i) $p(x\leq3\cup x>7$. i use one of the formula: $p(x\leq3\cup x>7)=p(x\leq3) + p(x>7)-p(x\leq3\cap x>7)$

$p(x\leq3\cap x>7)$=0, because of it's not possible? the probability that x is less than or equal to 3, $x=[0,1,2,3]$ and x is greater than 7, $(8,\infty)$

is this correct thought?

Answer 1

Let $A=\{x:x\leq3,x \in N\},B=\{x:x>7,x \in N\}$ since $A\cap B= O$ then $P(A \cup B) = P(A) +P(B)$ so yes u are right.