How to find the probability of taking a certain ball from a box after taking another one?

Question

Given a box with m balls. a is the number of black balls, b is the number of white balls, and a+b=m.

One random ball is taken from the box. What is the probability to take a white ball?

The correct answer is b/m, but I do not understand why.

Answer 1

With replacement, taking a ball and putting it back without looking at it makes no difference to the probability of the next

Without replacement

• the first ball might be black with probability $\frac a m$ in which case the probability of the next ball being white is $\frac{b}{m-1}$

• the first ball might be white with probability $\frac b m$ in which case the probability of the next ball being white is $\frac{b-1}{m-1}$

making the overall probability of the second ball being white without seeing the first being $$\frac a m \times \frac{b}{m-1}+ \frac b m \times\frac{b-1}{m-1} = \frac{ab +b(b-1)}{m(m-1)} = \frac{b(a+b-1)}{m(m-1)}= \frac{b}{m}$$