Given this question
Suppose urn #1 has 3 red and 2 blue balls, and urn #2 has 4 red and 7 blue balls. Suppose one of the two urns is selected with probability 1/2 each, and then one of the balls within that urn is picked uniformly at random. What is the probability that urn #2 is selected at the first stage (event A) and a blue ball is selected at the second stage (event B)?
And according to this book, Page 22 (Page 35 in PDF), P(B | A) is $\dfrac{7}{11}$
Why is it just $\dfrac{7}{11}$?
I understand it is because there is 4+7 balls in total and 7 of them are blue but I don't understand why it is not necessary to consider urn #1(the A part), that is, why not $\dfrac{7}{11}$ + $\dfrac{2}{5}$ or something to do with A
$P(B|A)$ is the probability of the occurrence of event $B$ given that event $A$ has already taken place. So, you know that urn 2 has been chosen. Hence, $$P(B|A)=P(\text{blue ball is picked from urn 2})=\frac{7}{11}.$$