# Joint and conditional with sets

I have a universe, $$U = \{a, b, c, d, e, f\}$$ and sets $$A = \{a, b, c\} and B = \{a, d, e, f\}$$

If $$P(A) = P(X = x \in A)$$ and $$P(B) = P(X = x \in B)$$, where $$X$$ is a random variable defined by uniformly selecting elements of $$U$$.

Are these values of unconditional, joint and conditional probabilities correct?

$$p(A) = 1/2$$

$$p(A,B) = 1/6$$

$$P(A|B) = 1/4$$

If you are assuming that each object of your universe has the same probability then each object has $$P(x=x)=\frac{1}{6}$$. So as $$a,b$$ and $$c$$ have the same probability and we are assuming they are independent, then $$P(x\in A) = P(x=a)+P(x=b)+P(x=c) = \frac{3}{6} = \frac{1}{2}$$. Then, $$P(x \in A \cap B) = P(x = a) = \frac{1}{6}$$ because $$A \cap B = a$$. Finally, $$P(A|B)$$ is the probability of $$a$$ in $$B$$ universe, so here, as ther are 4 objects with the same probability then each one has $$P(x=x)=\frac{1}{4}$$, so $$P(x \in A)=P(x \in A\cap B)=P(x=a) = \frac{1}{4}$$.
Another way is using the conditional formula $$P(A|B)=\frac{P(A\cap B)}{P(B)} = \frac{P(x=a)}{P(x\in B)} = \frac{\frac{1}{6}}{\frac{4}{6}} = \frac{1}{4}$$. So yes, your answer where correct.