# A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Calculating Probabilities of Repeated Independent Events, Part 1

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

In the remainder of Chapter IV of his Calculus of Probabilities, Markov proceeded to analyze the binomial distribution (as we would now call it), first covered in Chapter II, which can be viewed as the sum of independent Bernoulli random variables. He made use of some rather sophisticated arguments involving continued fractions, Stirling's approximations and hypergeometric series. This is definitely more advanced than the previous material.

In conclusion of the chapter, we return to an important question about the repetition of independent experiments that we studied in the second chapter.

Denoting the number of experiments by the letter $n$ and assuming that, for each of them, the probability of event $E$ is equal to $p$, we found that the probability that event $E$ occurs exactly $m$ times in these $n$ experiments is expressed by the product $\frac{1\cdot 2\cdot 3\cdot \cdots \cdot n}{1\cdot 2\cdot 3\cdot \cdots \cdot m \cdot 1\cdot 2\cdot 3\cdot \cdots \cdot (n-m)} p^m q^{n-m},$ where $q = 1-p$.

Therefore, the probability that event $E$ occurs more than $l$ times in the $n$ experiments considered is represented by the sum
$\frac{1\cdot 2\cdot 3\cdot \cdots \cdot n p^{l+1}q^{n-l-1}}{1\cdot 2\cdot 3\cdot \cdots \cdot (l+1) \cdot 1\cdot 2\cdot 3\cdot \cdots \cdot (n-l-1)} + \frac{1\cdot 2\cdot 3\cdot \cdots \cdot n p^{l+2}q^{n-l-2}}{1\cdot 2\cdot 3\cdot \cdots \cdot (l+2) \cdot 1\cdot 2\cdot 3\cdot \cdots \cdot (n-l-2)} + \cdots,$ which reduces to the product of the expression $P =\frac{1\cdot 2\cdot 3\cdot \cdots \cdot n}{1\cdot 2\cdot 3\cdot \cdots \cdot (l+1) \cdot 1\cdot 2\cdot 3\cdot \cdots \cdot (n-l-1)} p^{l+1}q^{n-l-1}$ and the sum $S = 1+ \frac{n-l-1}{l+2} \frac{p}{q} + \frac{(n-l-1)(n-l-2)}{(l+2)(l+3)} \left( \frac{p}{q} \right)^2 + \cdots$

For the approximate calculation of $P$ for large values of $n$, $l+1$ and $n-l-1$, we can use Stirling's Formula,25 which provides a series of inequalities, of which we show here only two of the simplest: $P < P_1 = \sqrt{\frac{n}{2\pi (l+1)(n-l-1)}}\left(\frac{np}{l+1}\right)^{l+1} \left(\frac{nq}{n-l-1}\right)^{n-l-1}$ and $\frac{P}{P_1} > H = e^{\frac{1}{12n} - \frac{1}{12(l+1)} - \frac{1}{12(n-l-1)}}.$

Continue to the second part of Markov’s analysis of the binomial distribution.