A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 2 – Solution

Review statement of Problem 2.

Solution: This problem can be considered as a special case of the previous, with \(a = \alpha\).

Namely, there are \(i\) tickets, whose numbers are chosen in advance, that are similar to the white balls, and the remaining tickets are similar to the black balls.

This similarity immediately reveals that the solution to the posed problem is obtained from the previous through substituting all numbers \(a,\, b,\, \alpha,\, \beta\) by the corresponding numbers \(i,\, n-i,\, i,\, m-i\).

Returning on that basis to the expression \[\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot (\alpha + \beta)}{1\cdot 2 \cdot\cdots \cdot \alpha \cdot 1\cdot 2 \cdot\cdots \cdot \beta} \cdot \frac{a(a-1) \cdots(a- \alpha + 1) \cdot b(b-1) \cdots (b- \beta + 1)}{(a+b)(a+b - 1) \cdots (a + b - \alpha - \beta +1)}\] found earlier, and making the indicated substitutions, we find the value of the desired probability in the form of the product \[\frac{1\cdot 2 \cdot 3 \cdot\cdots \cdot m}{1\cdot 2 \cdot\cdots \cdot i \cdot 1\cdot 2 \cdot\cdots \cdot (m-i)} \, \frac{i(i-1) \cdots 1 \cdot (n-i)(n-i-1) \cdots (n-m + 1)}{n(n - 1) \cdots (n-m +1)}\] which, after cancellation, reduces to⁶ \[ \frac{m (m-1) \cdots (m-i+1)}{n(n-1) \cdots (n-i+1)}.\]

Thus, the desired probability that among the chosen \(m\) numbers all of the \(i\) numbers indicated in advance appear is expressed by the fraction \[\frac{m (m-1) \cdots (m-i+1)}{n(n-1) \cdots (n-i+1)}.\]

Here, as with the first problem, Markov includes a second solution, which we omit.

Continue to an application of Problem 2.

Skip to statement of Problem 3.