Review statement of Problem 1.

*Solution*: Assume that all the balls in the vessel are distinguished from one another by numbers in such a way that the white balls have numbers \(1, 2, 3, \dots, a\) and the black have numbers \(a+1, a+2, \dots, a+b\).

The numbers on the chosen balls must form some set of \(\alpha + \beta\) numbers from all \(a+b\) numbers \(1, 2, 3, \dots, a+b\).

The number of different sets of \(\alpha + \beta\) numbers that can be formed from \(a+b\) numbers is equal to \[\frac{(a+b)(a+b-1)(a+b-2)\cdots (a + b - \alpha - \beta +1)}{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}.\]

Corresponding to this, we can distinguish \[\frac{(a+b)(a+b-1)(a+b - 2) \cdots (a+b - \alpha - \beta +1)}{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}\] equally likely cases, each of which consists of the appearance of a specific \(\alpha + \beta\) numbers.

From all these exhaustive and disjoint cases, the appearance of \(\alpha\) white and \(\beta\) black balls is favored only for those in which any set of \(\alpha\) numbers from the group \(1, 2, 3,\dots, a\), together with any \(\beta\) numbers from the group \(a+1, a+2, \dots, a+b\), appear.

The number of distinct sets of \(\alpha\) numbers that can be formed from \(a\) numbers is equal to \(\frac{a(a-1) \cdots (a- \alpha + 1)}{1\cdot 2\cdot\cdots \cdot \alpha}\) and the number of distinct sets of \(\beta\) numbers that can be formed from \(b\) numbers is equal to \(\frac{b(b-1) \cdots (b- \beta + 1)}{1\cdot 2\cdot\cdots \cdot \beta}\).

Therefore, the number of distinct sets of \(\alpha + \beta\) numbers that can be formed from \(a+b\) numbers is expressed by the product \[ \frac{a(a-1) \cdots (a- \alpha + 1)}{1\cdot 2\cdot\cdots \cdot \alpha} \cdot \frac{b(b-1) \cdots (b- \beta + 1)}{1\cdot 2\cdot\cdots \cdot \beta}.\]

Then the number of cases considered that favor the appearance of \(\alpha\) white and \(\beta\) black balls is expressed by the indicated product.

And, consequently, the desired probability that among the chosen \(\alpha + \beta\) balls, there will be \(\alpha\) white and \(\beta\) black is expressed by the ratio \[\frac{\frac{a(a-1) \cdots (a- \alpha + 1)}{1\cdot 2\cdot\cdots \cdot \alpha} \cdot \frac{b(b-1) \cdots (b- \beta + 1)}{1\cdot 2\cdot\cdots \cdot \beta}} {\frac{(a+b)(a+b-1)(a+b-2)\cdots (a + b - \alpha - \beta +1)}{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}},\] which, after simple transformations, reduces to^{4} \[\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}{1\cdot 2 \cdot\cdots \cdot \alpha \cdot 1\cdot 2 \cdot\cdots \cdot \beta} \cdot \frac{a(a-1) \cdots(a- \alpha + 1) \cdot b(b-1) \cdots (b- \beta + 1)}{(a+b)(a+b - 1)(a+b - 2) \cdots (a + b - \alpha - \beta +1)}.\]

Continue to Markov's presentation of a numerical example for Problem 1.

Skip to statement of Problem 2.

[4] In more compact notation, the solution reduces to: \(\frac{\binom{a}{\alpha} \binom{b}{\beta}}{\binom{a+b}{\alpha+\beta}}\).