I am trying to find upper bound of a conditional probability, conditioned on a low probability event. Consider the following scenario. Let $X$ and $Y$ be i.i.d. standard normals, such that $X+Y$ is distributed as $N(0,2)$. Let $a$ and $b$ be two positive reals, such that $a<b$. If $a$ and $b$ are large enough, then $P(a\leq X+Y\leq b)$ has a small probability. I am interested in finding out $P(X<0|a\leq X+Y\leq b)$.
I was wondering if I am on the right track to find such an upper bound or if there is a better way.
$\bf{Case 1:}$ Consider a simpler version of the problem, where we want to find $P(X<0|a\leq X+Y)$.
My approach is to write:
\begin{equation}
P(X<0|a\leq X+Y)=\frac{P(X<0, a\leq X+Y)}{P(a\leq X+Y)}
\end{equation}
For the numerator, $X+Y\geq a\Rightarrow Y\geq a-X$. Now, $X<0\Rightarrow a<a-X$. Combining these two, we get $Y>a$, implying, $P(X<0, a\leq X+Y)\leq P(Y>a)$. Therefore, $P(X<0|a\leq X+Y)\leq\frac{P(Y>a)}{P(X+Y\geq a)}$.
Is my approach correct and is this a reasonable upper-bound for the conditional probability? Is it possible to get a better upper bound using basic probability rules?
$\bf{Case 2:}$ Now for $Pr(X<0 | a\leq X+Y\leq b)$ we write in a similar fashion \begin{equation} P(X<0|a\leq X+Y\leq b)=\frac{P(X<0, a\leq X+Y\leq b)}{P(a\leq X+Y\leq b)} \end{equation}
and can get a bound $P(X<0|a\leq X+Y\leq b)\leq \frac{P(y>a)}{P(a\leq X+Y\leq b)}$
Can we get a better upper-bound?