# A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 1 – Solution

Author(s):
Alan Levine (Franklin and Marshall College)

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 to4 $\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}{1\cdot 2 \cdot\cdots \cdot \alpha \cdot 1\cdot 2 \cdot\cdots \cdot \beta} \cdot \frac{a(a-1) \cdots(a- \alpha + 1) \cdot b(b-1) \cdots (b- \beta + 1)}{(a+b)(a+b - 1)(a+b - 2) \cdots (a + b - \alpha - \beta +1)}.$

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

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