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

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.

Skip to statement of Problem 2.