70. a) From a box containing $n$ white and $m$ black balls, $k$ balls are drawn at random. What is the probability that exactly $r$ of the drawn balls will be white?

b) Apply the result of part a) to find the value of the sum

$$
C_{n}^{0} C_{m}^{k}+C_{n}^{1} C_{m}^{k-1}+C_{n}^{2} C_{m}^{k-2}+\ldots+C_{n}^{k} C_{m}^{0}
$$

Note. Other methods for finding this sum are discussed in the solutions to problems $55 m), 58$ ), 59 v).
70. a) Here, we consider an experiment consisting of drawing $k$ balls from a box containing $n+m$ balls. Choosing any $k$ balls from $n+m$ can be done in $C_{n+m}^{k}$ different ways. This is the number of equally probable outcomes of our experiment. Favorable outcomes will be those in which exactly $r$ white balls and, consequently, $k-r$ black balls are among the $k$ drawn balls. Clearly, for this to happen, it must be that $r \leqslant k, r \leqslant n$, and $k-r \leqslant m$; otherwise, there will be no favorable outcomes, and the probability will be zero. If these inequalities are satisfied, the number of favorable outcomes can be obtained by combining $C_{n}^{r}$ ways of drawing $r$ white balls from $n$ available white balls with $C_{m}^{k-r}$ ways of drawing $k-r$ black balls from $m$ such balls. Thus, the number
1) This answer initially seems paradoxical, as $\frac{3}{4}>\frac{5}{8}$, and it seems that winning three out of four games against an equally skilled opponent should be more difficult than winning five out of eight. However, this reasoning does not take into account that with an increase in the total number of games, the probability of a specific outcome generally decreases significantly due to the increase in the number of possibilities. In reality, one must distinguish between the probability of winning exactly five out of eight games, which was discussed in the problem, and the probability of winning at least five out of eight games. It is indeed easier to win at least five out of eight games against an equally skilled opponent than to win at least three out of four, but for winning at least five out of eight games, there are four different possibilities (winning five, six, seven, or eight games), while for winning at least three out of four games, there are only two (winning three or four games).
of favorable outcomes is $C_{n}^{r} C_{m}^{k-r}$, and the desired probability is $\frac{C_{n}^{r} C_{m}^{k-r}}{C_{n+m}^{k}}$.

b) Let $k \leqslant n$ and $k \leqslant m$. When drawing $k$ balls from a box containing $n$ white and $m$ black balls, the number of white balls drawn can be 0, 1, 2, ..., or $k$. According to part a) of the problem, the probability that this number is 0 is $\frac{C_{n}^{0} C_{m}^{k}}{C_{n+m}^{k}}$; the probability that it is 1 is $\frac{C_{n}^{1} C_{m}^{k-1}}{C_{n+m}^{k}}$; the probability that it is 2 is $\frac{C_{n}^{2} C_{m}^{k-2}}{C_{n+m}^{k}}; \ldots$; the probability that it is $k$ is $\frac{C_{n}^{k} C_{m}^{0}}{C_{n+m}^{k}}$. The sum of all these probabilities must equal one: indeed, adding them all together, we get a fraction where both the numerator and the denominator represent the total number of equally probable outcomes of the experiment. Therefore,

$$
\frac{C_{n}^{0} C_{m}^{k}}{C_{n+m}^{k}}+\frac{C_{n}^{1} C_{m}^{k-1}}{C_{n+m}^{k}}+\frac{C_{n}^{2} C_{m}^{k-2}}{C_{n+m}^{k}}+\ldots+\frac{C_{n}^{k} C_{m}^{0}}{C_{n+m}^{k}}=1
$$

and, consequently,

$$
C_{n}^{0} C_{m}^{k}+C_{n}^{1} C_{m}^{k-1}+C_{n}^{2} C_{m}^{k-2}+\ldots+C_{n}^{k} C_{m}^{0}=C_{n+m}^{k}
$$

Note. The method used in problem 70b) to derive relations involving binomial coefficients is conceptually similar to the "geometric" method discussed in problem 59. In both cases, we partition a certain set of objects (shortest paths in the "geometric" method and all possible outcomes of the experiment in the solution to problem 70b)) into disjoint subsets and then equate the total number of our objects to the sum of the numbers of objects in the subsets. Note that the total number of experimental outcomes, unlike the number of shortest paths, does not necessarily express itself as a binomial coefficient, which makes the method of problem 70b) applicable to deriving a series of relations that cannot be obtained from the geometric scheme of problem 59 (see, for example, problem 71).