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Say you have two random continuous variables $X$ and $Y$, and their joint probability distribution is $f(x,y)$.

You want to compute

$$P(a < X < b | Y = y)$$

If using the definition of conditional probability you get

$$P(a < X < b | Y = y) = \frac{P((a < X < b)\cap Y = y)}{P(Y=y)}$$

But $P(Y = y)$ is by definition $0$ because a continuous random variable cannot assume one value. So how is conditional probability defined in this case?

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You are supplied with a joint density function, so first express the probability as an integral, then apply the definition of conditional probability density.

$$\mathsf P(a< X< b\mid Y=y) ~{~=~\int_a^b f_{X\mid Y}(s\mid y)~\mathrm d s\\ ~=~\int_{\infty}^\infty \frac{f_{X,Y}(s, y)}{f_Y(y)}~\mathrm d s\\ ~=~ \dfrac{\int_{a}^b f_{X, Y}(s,y)~\mathrm d s}{\int_{-\infty}^\infty f_{X, Y}(s,y)~\mathrm d s}\\ ~=~ \dfrac{\int_{a}^b f(s,y)~\mathrm d s}{\int_{-\infty}^\infty f(s,y)~\mathrm d s}}$$

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In the multivariate case $P(Y=y)$ must not be $0$. The graph below shows a joint distribution of two normals on approximately $-5\leq X,Y \leq 5$.

For $Y=-5$ you can see a marked area. This area looks like the representation of a density of the univariate case. If someone looks at the area nobody would come in mind that $P(-2< X < 2)=0$, for instance.

Therefore $P(Y=-5)$ is greater than $0$.

enter image description here

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