# A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Calculating Probabilities of Repeated Independent Events, Part 3

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

It remains to compare the previous inequalities with those which $$P$$ satisfies and which we have shown above, and we will have the possibility of forming a series of approximate values of the probability that event $$E$$ occurs more than $$l$$ times in the $$n$$ experiments considered, where for each of these approximate values, we will know whether it is greater than the probability or, conversely, less than it.28

Based on these inequalities, interchanging $$p$$ with $$q$$ and substituting $$l$$ by $$n-l'$$, we find a series of approximate values of the probability that event $$E$$ occurs less than $$l'$$ times in the $$n$$ considered experiments, where for each of the given approximate values of this new probability, we will also know whether it is larger than the probability or, conversely, smaller than it.

And with approximate values of the probability that event $$E$$ occurs more than $$l$$ times and the probability that event $$E$$ occurs less than $$l'$$ times, where $$l > l'$$, it is not difficult to obtain an approximate value of the probability that event $$E$$ occurs not more than $$l$$ times and not less than $$l'$$ times, since the sum of these three probabilities must be 1.

For example, let $$p=\frac{3}{5}$$, $$q=\frac{2}{5}$$, $$n=6520$$ and we will seek the probability that the ratio of the number of occurrences of event $$E$$ to the number of experiments will differ from $$\frac{3}{5}$$ by less than $$\frac{1}{50}$$.

In other words, we will seek the probability that event $$E$$ occurs not more than 4042 times, and the complement to the event not more than 2738 times.29

According only to the established remarks, calculation of the desired probability reduces to calculating the probability that event $$E$$ occurs more than 4042 times and the probability that the complementary event occurs more than 2738 times.

Returning to the probability that event $$E$$ occurs more than 4042 times, in the above formulas and inequalities, we must let $$p=\frac{3}{5}$$, $$q= \frac{2}{5}$$, $$n=6520$$, $$l = 4042$$.

Then: $P_1 = \sqrt{ \frac{3260}{\pi\cdot 4043 \cdot 2477}} \left(\frac{3912}{4043}\right)^{4043} \left(\frac{2608}{2477}\right)^{2477},$ $H = e^{\frac{1}{12\cdot 6520} - \frac{1}{12\cdot 4043} - \frac{1}{12\cdot 2477}},$ and by means of logarithmic tables, we find...

The rest of this page in the original text is filled with highly-detailed arithmetic involving logarithms with 10-digit mantissas. We shall omit this and jump to Markov's conclusions.

$\log H > -0.00003,$ $0.00004094 < P < 0.00004095.$

On the other hand, we have:
$c_1 = \frac{2477}{4044} \cdot \frac{3}{2} = \frac{7431}{8088},\ \ d_1 = \frac{6521}{4044\cdot 4045} \cdot \frac{3}{2} = \frac{19563}{32715960},$
$c_2 = \frac{2476 \cdot 4044}{4045 \cdot 4046}\cdot \frac{3}{2} = \frac{7509708}{8183035},\ \ d_2 = \frac{6522 \cdot 3}{4046 \cdot4047} = \frac{3261}{2729027},$
$c_3 = \frac{2475 \cdot 4045}{4047 \cdot 4048} \cdot \frac{3}{2} = \left(1 - \frac{2}{4047} \right) \frac{7425}{8096}.$
and, by performing simple calculations we get, consecutively,
$c_3< 0.9167,\ \ \frac{d_2}{1-\omega_3} < \frac{3261}{0.0833\times 2729027} < 0.01435,$
$0.918 > c_2 > \omega_2 >\frac{c_2}{1.01435} > 0.9047,$
$0.0074 > \frac{d_1}{0.082} > \frac{d_1}{1-\omega_2} > \frac{d_1}{0.0953} > 0.00626,$
$0.912 < \frac{c_2}{1.0074} < \omega_1 < \frac{c_1}{1.00626} < 0.9131,$
$11.36 < \frac{1}{0.088} < S < \frac{1}{0.0869} < 11.51;$
Consequently,
$SP < \frac{0.4095}{869} < 0.0004713,$ but
$SP > \frac{0.4094}{880} > 0.000465.$

Proceeding to the probability that the event complementary to $$E$$ occurs more than 2738 times, we must let $$p=\frac{3}{5}$$, $$q= \frac{2}{5}$$, $$n=6520$$, $$l = 2738$$, ...

This page is similar to the previous pages in featuring detailed arithmetic using logarithms.

$\log H < -0.00003,$ $0.00004280 < P < 0.00004281,$ $\dots$ $SP < 0.0005002,$ $SP > 0.000491.$

So, the probability that event $$E$$ occurs more than 4042 times in the 6520 considered experiments is included between 0.0004713 and 0.000465, and the probability that in these experiments, event $$E$$ occurs less than 3782 times is included between 0.0005002 and 0.000491.

And, therefore, the probability that event $$E$$ occurs not less than 3782 times and not more than 4042 times lies between $$1 - 0.000972 = 0.999028$$ and $$1 - 0.000956 = 0.999044$$.

Continue to Suggestions for the Classroom and Conclusion.

[28] In essence, Markov was computing the “convergents” of the continued fraction, where by the $$k$$th convergent $$c_k$$ of an irrational number $$r$$, we mean the rational number obtained by truncating the continued fraction to the first $$k$$ terms. It can be shown that $c_0 < c_2< c_4< \cdots < r< \cdots< c_5< c_3<c_1,$ so the convergents alternate between being less than $$r$$ and greater than $$r$$. Both sides converge to $$r$$. Furthermore, if we express $$c_k = \frac{p_k}{q_k}$$ as a fraction in lowest terms, there is no rational number with denominator less than $$q_k$$ that is closer to $$r$$.

[29] $$4042 = 6520\cdot (\frac{3}{5} + \frac{1}{50})$$; while $$2738 = 6520- 6520\cdot (\frac{3}{5}-\frac{1}{50})$$.