Let P be a measure of probability in a sample space S. Can there be events A and B such that $P (A | B) = \frac{1}{3}$, $P (B | A) = \frac{1}{10}$ and $P (B) =\frac{1}{2}$ ?
My attempt has been to make a system of equations with the conditioned probability equation and see if this has a solution, but I don't know how to interpret the condition of the intersection. I would appreciate your help, I'm new to probability. Thank you
Bayes theorem would given you $P(B|A) = \frac{P(A|B)P(B)}{P(A)}$
$\therefore P(A) = \frac{1/6}{1/10} = \frac{10}{6} > 1$ which is a contradiction.
So, such a scenario cannot occur.