# The probability that the second ball is white, without knowing type of the first drawn ball

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 ?

There are $$M-1$$ balls left. So what?