Bound in the probability of union when only a fixed number of events happen Given $\{X_{i}\}_{i=1}^{n}$ iid continous random variable and we know for a fact that always (if you want to be precise almost surley) we have $\displaystyle{\sum_{i=1}^{n}}\mathbf{1}_{X_{i} \in A} = k$ then I'm wondering if the following is true
\begin{align*}
    \mathbb{P}\left(\displaystyle{\bigcup_{i=1}^{n}}X_{i} \in A, X_{i} \leq t\right) &\leq k \mathbb{P}(X_{1} \in A, X_{1} \leq t)
\end{align*}
I know the result it's true if we replace $k$ for $n$ in the upper bound but I'm wondering if we can do better since we know all the events can't happen. I really have no clue where to start a counterexample or a proof so any help would be welcome.
Edit: Just for a bit of context on A, in my case A could be something like $X_i \in A \iff X_i \text{ is less or equal than the k-th statistics}$
 A: The simplest way to get a counterexample is to take $\ A\ $ to be bounded above, and $\ t\ge\sup A\ $. Then
$$
\big\{X_i\in A, X_i\le t\big\}=\big\{X_i\in A\big\}
$$
and if $\ k>0\ $, then
$$
\left\{\sum_{i=1}^n \mathbf{1}_{X_i\in A}=k\right\}\subseteq\bigcup_{i=1}^n\big\{X_i\in A\big\}=\bigcup_{i=1}^n\big\{X_i\in A, X_i\le t\big\}\ .
$$
Therefore
\begin{align}
\mathbb{P}\left(\bigcup_{i=1}^n\big\{X_i\in A, X_i\le t\big\}\right.&\,\left|
\,\sum_{i=1}^n \mathbf{1}_{X_i\in A}=k\right)\\
&=\mathbb{P}\left(\bigcup_{i=1}^n\big\{X_i\in A\big\}\,\left|
\,\sum_{i=1}^n \mathbf{1}_{X_i\in A}=k\right.\right)\\
&=1\ .
\end{align}
Therefore, to get a counterexample, we simply need to choose $\ X_i\ $ and $\ A\ $ so that $\ \mathbb{P}\big(X_i\in A\big)<\frac{1}{k}\ $. Then
\begin{align}
\mathbb{P}\left(\bigcup_{i=1}^n\big\{X_i\in A, X_i\le t\big\}\right.\,\left|
\,\sum_{i=1}^n \mathbf{1}_{X_i\in A}=k\right)&=1\\
&>k\mathbb{P}\big(X_i\in A\big)\\
&=k\mathbb{P}\big(X_i\in A, X_i\le t\big)\ .
\end{align}
If you choose the distribution of the $\ X_i\ $ to be continuous—normal, say, or uniform over an interval of positive length, for example—then the probabilities in this inequality will be continuous functions of $\ t\ $. You could therefore then decrease the value of $\ t\ $ to just below $\ \sup A\ $ by a small amount and the inequality would still hold.
