Conditional probability confusion I have a continuous random variable X and I am told to find the following probability:
$ P(X\ge 3 | X \le 4)$
I understand that:
$ P(A|B)=\frac{P(A \cap B)}{P(B)} $
However, in my head it seems logical to say:
$ P(X\ge 3 | X \le 4) = P(3 \le X \le 4)$
Why is this not correct?
Thanks,
Matt
 A: A numeric example of why it is wrong:
Roll a fair $6$-sided die, and let $X$ denote the result:


*

*$P(X\geq3|X\leq4)=\text{the probability of getting $3$ or $4$ from $[1,2,3,4]$}=\frac24$

*$P(3\leq{X}\leq4)=\text{the probability of getting $3$ or $4$ from $[1,2,3,4,5,6]$}=\frac26$
A: Let's say we're rolling a die, so our sample space is $\{1, 2, \ldots, 6\}$, and $P(X = x) = 1/6$.
Then $P(3 \leq x \leq 4)$ is just $2/6$, while $P(X \geq 3 \mid X \leq 4)$ would be  $2/4$. In this second scenario, once we're given that $X \leq 4$, then it's essentially like we're rolling a $4$-sided die, so we have a $1/4$ chance of getting any particular number, thus a $2/4$ probability of having $3 \leq X \leq 4$.
If we aren't given any information, like the fact that $X \leq 4$, then we're still rolling our standard $6$-sided die. So, the condition that $X \geq 4$ essentially limits our sample space; we're no longer looking at $X \in \{1, 2, \ldots 6\}$, but instead $X \in \{1, 2, 3, 4\}$. This smaller sample spaces (in this case) increased the probability that $3 \leq X \leq 4$.
A: If $\mathbb P(B)>0$ then the conditional probability of $A$ given $B$ is defined as
$$\mathbb P(A|B) = \frac{\mathbb P(A\cap B)}{\mathbb P(B)}. $$
In our example, say $X$ has density $f$, $A=\{X\geqslant3\}$, and $B=\{X\leqslant 4\}$. If $\mathbb P(B)=1$, i.e. 
$$\int_4^\infty f(x)\mathsf dx = 1, $$
then indeed $\mathbb P(A|B) = \mathbb P(A\cap B)$ is valid. If $\mathbb P(A\cap B)=0$, i.e.
$$\int_3^4 f(x)\mathsf dx = 0, $$
then again $\mathbb P(A|B)=\mathbb P(A\cap B)$ (as they are both zero). If $\mathbb P(A\cap B)>0$ and $\mathbb P(B)<1$, though, we have
$$\mathbb P(B) = \frac{\mathbb P(A\cap B)}{\mathbb P(A|B)}\ne1, $$
so that $\mathbb P(A|B)\ne \mathbb P(A\cap B)$, that is,
$$\mathbb P(X\geqslant 3|X\leqslant 4)\ne \mathbb P(3\leqslant X\leqslant 4). $$
