# Finding $P(X>1)$ and $P(X\geq 1)$ from CDF

Introduction Statistics questions. I hope my question isn't too basic for this platform. I am given the CDF

I've found $$P(X \leq 1)$$. I've found $$P(X > 1)$$ by $$1-P(X \leq 1)$$.

Now I'm asked for $$P(X\geq 1)$$.

Is this the same as $$P(X > 1)$$? I know that $$P(X=1)=0$$ so does $$P(X\geq 1)=P(X > 1)+P(X=1)$$?

I'm also asked to find, $$P(−1.5 < X < 0.5)$$. I've previously only seen problems of the form $$P(−1.5 < X \leq 0.5)$$ and know to solve them by subtraction.

Intuitively, it seems that by including the possibility that $$X=1$$ would make $$P(X\geq 1) > P(X > 1)$$ but since $$P(X=1)=0$$, I suppose it doesn't.

For the purposes of my homework, how do I treat problems that are inclusive or exclusive of the number being asked about?

For my curiosity, there most be some difference in practice, right? $$P(X\leq 1)$$ would be close to $$P(X\leq .99999999)$$ which would be different than $$P(x\leq 1)$$. Or am I overthinking things?

• Your idea seems OK. Distribution is continuous between -2 and 2. Sep 6 '20 at 15:44
• – whuber
Sep 6 '20 at 15:57

If $$A$$ and $$B$$ are disjoint, measurable sets, then $$P(A \cup B) = P(A) + P(B).$$ Disjoint means that the two sets don't overlap, i.e. the intersection is the empty set, or symbolically $$A \cap B = \emptyset$$. Measurable is a more technical concept, but (probably) every set you've ever seen is measurable, and any set like $$\{ X < a \}$$, $$\{ X \le a \}$$, $$\{ X = a \}$$, $$\{ X \ge a \}$$, or $$\{ X > a \}$$ is measurable.
This means that $$P( X \le a ) = P(\{ X < a \} \cup \{ X = a \}) = P(X < a) + P(X = a)$$ and $$P( X \ge a ) = P(\{ X > a \} \cup \{ X = a \}) = P(X > a) + P(X = a)$$
Since you've correctly noticed that because your cdf $$F$$ is continuous, $$P(X = a) = 0$$ for any number $$a \in \mathbb{R}$$, you can conclude that $$P(X \le a) = P(X < a) \quad \text{and} \quad P(X \ge a) = P(X > a).$$