# Is it true that $\mathbb P (\max(X,Y) \leq t \vert X \geq Y)= \mathbb P(X \leq t)$?

Sorry for the probably very easy question;

From a mathematical perspective I would have said yes, it holds because

$$\mathbb P (\max(X,Y) \leq t \vert X \geq Y)= \mathbb P(X \leq t \vert X \geq Y)$$

and now the condition about $$Y$$ becomes unimportant since it is not part of the probability anymore and can be omitted - however if I know that some random variable is bigger than another (or in a bigger example bigger than a bunch of random variables $$Y_1,...,Y_n$$), then the probability that $$X$$ is very big should also be higher or am I wrong?

Counterexample. Even if we imagine that $$X$$ and $$Y$$ are independent, suppose that $$Y$$ always equals $$1$$, $$X$$ equals $$0$$, $$2$$, or $$4$$ with probability $$1/3$$ each, and $$t = 3$$. Then
$$P(\max(X, Y) \leq t \mid X \geq Y) = \frac{P(\max(X, Y) \leq t, X \geq Y)}{P(X \geq Y)} = \frac{1/3}{2/3} = \frac12$$
$$P(X \leq t) = \frac23$$