Struggling with intuition about this probability question. Symmetry argument of two balls drawn from an urn.

Question

So the question is as follows:

An urn contains m red balls and n blue balls. Two balls are drawn uniformly at random from the urn, without replacement.

(a) What is the probability that the first ball drawn is red?

(b) What is the probability that the second ball drawn is red?*

The answer to a) quite clearly works out to be $\frac{m}{(m+n)}$, but the answer to b turns out to be the same, and my tutor said this is intuitive by a symmetry argument.

i.e. that $P(A_1)$ = $P(A_2)$ where $A_i$ is the event that a red ball is drawn on the ith turn. However I am struggling to see how this is evident, can anyone explain this?

Answer 1

Part (b) doesn't give any information about the first ball, it is just asking for the probability that the second ball in the line is red.

Now red balls (or those of any other color!) don't have any preference for positions in the line, hence if you randomly pick up any ball from the line, its probability of being red will be the same.

Answer 2

The second probability is

$$\frac{m-1}{m+n-1},$$

which is not the same as the first, so there's nothing to be worried about.

Answer 3

The secon ball probability is$\frac{m}{m+n}\frac{m-1}{m+n-1}+\frac{n}{n+m}\frac{m}{m+n-1}=\frac{m}{n+m}$ This is the same as the first, which, by symmetry, is what it should be.