A die is thrown two times. $A$ is an event, where sum of obtained numbers is less than 6, and $B$ - that both numbers are even. What is $P(A\mid B)$?

Question

So, the probability, that event $B$ occurs is $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$. Probability that $A$ occurs is $\frac{1}{9}$. Probability that $B$ occurs when $A$ has occurred is $P(B \mid A)=\frac{1}{10}$. Then

$$P(A \mid B)=\frac{P(B\mid A) P(A)}{P(B)}=\frac{\frac{1}{10}\cdot \frac{5}{18}}{\frac{1}{4}}=\frac{20}{180}=\frac{1}{9}.$$

My question, is, first, if I need to show that these two events are not independent, do I do it just by showing that $P(A \mid B) \neq P(A)$? Then, I partly computed all the probability just by drawing a table, Because if I wanted to calculate $P(B \mid A)$, I would again need $P(A \mid B)$. Is there a way to calculate $P(B \mid A)$ in a more analytical way, than just by drawing? The same goes with the event $A$ itself, because it is not independent, so I just need to sum all the cases where the sums of the two dice are less than 6.

Answer 1

hint...it would be easier to use $$p(A\mid B) = \frac{p(A\cap B)}{p(B)}$$

Answer 2

You could show $P(A,B) \not=P(A)P(B)$ to show they are not independent

Once you have those three probabilities, i.e. $\frac{1}{36} \not = \frac{10}{36}\times\frac{9}{36}$, you can also immediately calculate $P(A \mid B) = \frac{P(A,B)}{P(B)} = \frac{1}{10}$ and $P(B \mid A) = \frac{P(A,B)}{P(A)} = \frac{1}{9}$ if you want to, but it is not necessary