Probability of A given B question

There are two bowls. The first bowl contains 3 white balls and 8 black balls. The second bowl contains 6 white balls and 5 black balls. To determine the bowl that you will chose, a fair six-sided dice will be rolled. If the resulting number is greater than 4, you will randomly select a ball from the first bowl, otherwise, you will randomly select a ball from the second bowl. What's probability that you chose from the first bowl given that a black ball was chosen.

So far I have the probability of choosing from first bowl is 2/6. Probability of choosing from second bowl is 4/6.

$$P(1|B)=\frac{P(B|1)P(1)}{P(B)}$$
As you said $P(1)=2/6$, $P(2)=4/6$
$P(B|1)=8/11$, $P(B|2)=5/11$
$P(B) = P(B|1)P(1)+P(B|2)P(2)$