There are $n$ urns of which the $i$-th urn contains $i − 1$ red balls and $n − i$ blue balls. You pick an urn at random and remove two balls at random without replacement. Find the probability that (a) the second ball is blue; (b) the second ball is blue given the first ball is blue.
This is a repeated question, and I found its solution here in Find probability of specific ball getting selected on second turn .
I am having some problems with the solution. According to the solution, the probability is $2/3$.
Consider $n=3$ . Then the first urn has $2$ blue balls, second has $1$ red and $1$ blue ball, and the third has only $2$ blue balls
Hence probability of choosing a blue ball the second time, given the first one is blue is : $1/3\cdot1$ [ Since $2$ blue balls, one already taken, the other one is obviously blue] $+ 1/3\cdot0$ [ There was just one blue ball, the other was red. The blue ball was picked, hence only red ball remains] $+ 1/3\cdot0$ [ No blue balls] . Hence the probability must be $1/3$.
What is the problem in my logic?