Consider two random variables and as seen in the following joint PMF table: \begin{array}{|c|c|c|c|} \hline &Y=2& Y=3& Y=4 \\ \hline X=1 &1/12& 1/24& 1/24 \\ \hline X=2 &1/6& 1/12& 1/8 \\ \hline X=3 &1/4& 1/8& 1/12 \\ \hline \end{array}
I calculate the conditional probability of $P_{|}(1|2)$ as follows:
$$P(X=1|Y=2) = \dfrac{P((X=1) \cap (Y=2))}{P(Y=2)} = \dfrac{P(X=1,Y=2)}{P(Y=2)} = \dfrac{P_{X,Y}(1,2)}{P_{Y}(2)} = \dfrac{1/12}{1/2} = \dfrac{1}{6}$$
The way this conditional probability is phrased confuses to me:
$$P_{|}(1)$$, where $$=\{{>3}\}$$
Given , then only $ = 4$ satisfies the condition. To evaluate $P_{|}(1)$:
$$P(X=1|Y=4) = \dfrac{P((X=1) \cap (Y=4))}{P(Y=4)} = \dfrac{P(X=1,Y=4)}{P(Y=4)} = \dfrac{P_{X,Y}(1,4)}{P_{Y}(4)} = \dfrac{1/24}{1/4} = \dfrac{1}{6}$$
This seems about right or am I missing something?
I am unsure as to why $P_{|}(1)$ only contains one integer and excludes the vertical condition line.
So to solve $P_{|}(2)$ given the same condition for $=\{{>3}\}$, then $(=2|=4)$?
Any feedback is highly appreciated. Thanks!