Assume we are given a joint distribution $P(X,Y)$ where $P(0,0)=0.1$, $P(0,1) = 0.4$, $P(1,0)=0.3$, and $P(1,1)=0.2$. The goal is to compute $P(X|Y=1)$.
Traditionally, solving a conditional probability problem $P(A|B)$ simplifies to $\frac{P(A,B)}{P(B)}$, but I'm unsure how to apply it to this case. In particular, I am unclear on what the probability $P(X,Y=1)$ means since $P(X)$ is a marginal probability.
To avoid this, I enumerated the different values $X$ takes on and plugged it into the original quantity to solve -- $P(X|Y=1)$:
- $P(X=0|Y=1) = 0.4/0.6 = 4/6$
- $P(X=1|Y=1) = 0.2/0.6 = 2/6$
This gives me the final answer of $P(X|Y=1) = [4/6, 2/6]$, but I'm not sure whether the answer should be multiple probabilities or a single probability.