In the book of Shiryaev, Probability, on page 26, it is given that
EXAMPLE 2. An urn contains M balls, m of which are "lucky." We ask for the probability that the second ball drawn is lucky (assuming that the result of the first draw is unknown, that a sample of size 2 is drawn without replacement, and that all outcomes are equally probable).Let A be the event that the first ball is lucky, B the event that the second is l lucky. Then [...] $$P(B)= m/M.$$ lt is interesting to observe that $P(A)$ is precisely $m/M$. Hence, when the nature of the first ball is unknown, it does not affect the probability that the second ball is lucky.
Even though I can verify the correctness of the derivation given for this result in the book, I think, there is still something missing;
Yes, since we have given no information about the result of the first drawn ball, one would expect that the probability of choosing a lucky ball should not be affected by its order. However, the experiment is done without replacement. This means compared to the first drawn ball, we have a piece of extra information about the state of the urn, namely that a ball has chosen and discarded afterwards, so we know that there are now $M-1$ balls inside the urn. But, the fact that $P(B) = m/M$ does not account for this, so how come ?