Conditioning on a Continuous Variable with a Discrete Value

Question

Textbook Question: \begin{equation} f(x,y) = \begin{cases} x + y & \text{if $0\le x \le1$, $0\le y \le1$ } \\ 0 & \text{otherwise} \end{cases} \end{equation}

What is $P(X < \frac{1}{4} \mid Y = \frac{1}{3})$?

[edit: added solution for more clarity]

I'm CAN SOLVE the question by applying $P(X \in A \mid Y=y) = \int_A f(_X\mid_Y) (x\mid y)dx$

Deriving (1) $$ (1) - f(_X\mid_Y) (x\mid y) = \frac{f(_X,_Y)(x,y)}{f_Y(y)} = \frac{x+y}{y+\frac{1}{2}} $$ $$ P(X < \frac{1}{4} \mid Y=\frac{1}{3}) = \int_0^\frac{1}{4} f(_X\mid_Y) (x\mid \frac{1}{3})dx $$ $$ = \int_0^\frac{1}{4}\frac{x+\frac{1}{3}}{\frac{1}{3}+\frac{1}{2}}dx$$ $$ = \frac{11}{80} $$

My question is Conceptual However:

1. if Y is a Continuous Random Variable $P(Y=y) = 0$; meaning $P(Y=\frac{1}{3}) = 0$. Hence Event $Y = \frac{1}{3}$ CANNOT occur.
2. hence in the above Question I can't conceptually appreciate how to condition on an Event i.e $Y = \frac{1}{3}$ that CANNOT occur.

Thanks in advance for the explanations.

p/s: I had read previous Questions: Conditional probability when conditioning on continuous-discrete variables & Conditional probability combining discrete and continuous random variables But couldn't figure a clear answer hence raise this Q.

Apologies if it's too noob a question; just starting Statistics.

Answer 1

For completeness answering my Question based on the comments provided.

if Y is a Continuous Random Variable $P(Y=y)=0$; meaning $P(Y=\frac{1}{3})=0$. Hence Event $Y=\frac{1}{3}$ CANNOT occur.

Probability 0 events can occur. From a frequentist perspective, they just occur at a rate of () o(n), where n is the number of times you repeat the experiment.

hence in the above Question I can't conceptually appreciate how to condition on an Event i.e $Y=\frac{1}{3}$ that CANNOT occur.

assuming $P(Y=\frac{1}{3}) > 0$ we can use the (∈∣=)=∫(∣)(∣). Given the Mathematical resolution of this is complicated better taken as a definition for entry level.