28. From a box containing $n$ white and $m$ black balls $(n \geqslant m)$, all balls are drawn one by one at random. Find the probability of the following events:

(a) there will be a moment when the number of white and black balls in the sequence of drawn balls will be equal

(b) at any moment, the number of white balls in the sequence of drawn balls will not be less than the number of black balls.
Solution. (a) The number of all sequences is $C_{n+m}^{n}$. If a sequence starts with a black ball, it is favorable. There are $C_{n+m-1}^{m-1}$ such sequences. Between the favorable sequences $y$, the first ball of which is white, and the sequences $x$, starting with a black ball, there is a bijection $\varphi$. It is defined as follows: $\varphi(x)$ is the sequence obtained by mirror-reversing the balls from 1 to $t$ in the sequences $x$; here $t$ is the first moment when the number of white and black balls in the sequences $x$ will be equal. Therefore, the desired probability is

$$
\frac{2 C_{n+m-1}^{n-1}}{C_{n+m}^{n}}=\frac{2 m}{n+m}
$$

(b) The number of favorable sequences coincides with the number of paths from problem I.1.4. Therefore, the desired probability is

$$
\frac{C_{n+m}^{n}-C_{n+m}^{n+1}}{C_{n+m}^{n}}=\frac{n-m+1}{n+1}
$$

## § 3. Conditional Probabilities. Independence