Calculating mutual events with continuous distribution

Question

Suppose $\ X \sim U(-4,7) $

Compute: $\ P\{X^2 - 16 > 0 \ | \ X > 0 \} $

My attempt:

$$\ P\{ X^2 - 16 > 0 | X>0 \} = P \{ X > 4 | X > 0 \} = \frac{P\{X>4\}}{P\{X>0\} } $$

I'm just not sure I understand what how to calculate the probability of $\ P\{ X > a \} \cap P\{X > b\} $ with continues variables. Is it equal to $\ P\{X > a \} \cdot P\{X >b \} $

When I draw a uniform distribution it's seems like it is $\ P\{ X > a | X >b \} = \frac{P\{X>a\}}{P\{X>b\}} $ where $\ a > b $ but I'm not sure.

Answer 1

First verify that $\{X^2-16 > 0\} = \{X > 4\} \cup \{X < -4\}$.

Then by definition of conditional probability we have \begin{align} \mathbb{P}\big(\{X^2 - 16 > 0\} \mid \{X>0\}\big) &= \frac{\mathbb{P}(\{X^2-16 > 0\} \cap \{X>0\})}{\mathbb{P}(X>0)} \\ &= \frac{\mathbb{P}\big(\left(\{X > 4\} \cup \{X < -4\}\big) \cap \{X>0\}\right)}{\mathbb{P}(X>0)}\\ &= \frac{\mathbb{P}\big(\left(\{X > 4\} \cap \{X > 0\}\right) \cup \left(\{X <- 4\} \cap \{X > 0\}\right)\big)}{\mathbb{P}(X>0)}\\ &= \frac{\mathbb{P}\big(\{X > 4\}\cup \emptyset\big)}{\mathbb{P}(X>0)}\\ &= \frac{\mathbb{P}(X > 4)}{\mathbb{P}(X>0)}\\ &= \frac37 \end{align}

Answer 2

Well what is the intersection of the sets? What does it mean if $\omega$ is fulfilling $X(\omega) > a$ and $X(\omega) > b$?