Basic Probability proof question

Question

I have to proof the following statement: Show that there cannot exist three events $A, B, C$ satisfying $P(A ∩ B) = 1/8$

and

$P(B|A) = 1/2P(A)=1/3P(B)$

Hint try to compute $P(A\cup B)$

Why is event $C$ specified or is this just a typo? If so how to proof this (without event $C$)?

Answer 1

Note: $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac18}{P(A)}=\frac12 P(A) \Rightarrow P(A)=\frac12;\\ \frac12 P(A)=\frac13P(B) \Rightarrow P(B)=\frac34;\\ P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac12+\frac34-\frac18=\frac98>1.$$ So, it does not matter whether $C$ is given or not.

Answer 2

I think event $C$ is probably a typo, although the question still technically makes sense with it. You can show that the $A$ and $B$ described are impossible. And then it will also be the case that $A,$ $B$ and some unspecified and unimportant $C$ are impossible.

You have enough information to calculate $P(A)$ and $P(B)$ and then you can use the hint to show there's a contradiction.