Review statement of Problem 1.

\(a=3,\ b=4,\ \alpha =2,\ \beta= 2 \).

Let us assume the white balls are set by the numbers 1, 2, 3 and the black by the numbers 4, 5, 6, 7.

The numbers on the chosen four balls can be represented by any of the following \(\frac{7 \cdot 6 \cdot 5 \cdot 4}{1\cdot 2 \cdot 3 \cdot4} = 35 \) sets:

1, 2, 3, 4 | 1, 2, 3, 5 | 1, 2, 3, 6 | 1, 2, 3, 7 | 1, 2, 4, 5 | 1, 2, 4, 6 | 1, 2, 4, 7 |

1, 2, 5, 6 | 1, 2, 5, 7 | 1, 2, 6, 7 | 1, 3, 4, 5 | 1, 3, 4, 6 | 1, 3, 4, 7 | 1, 3, 5, 6 |

1, 3, 5, 7 | 1, 3, 6, 7 | 1, 4, 5, 6 | 1, 4, 5, 7 | 1, 4, 6, 7 | 1, 5, 6, 7 | 2, 3, 4, 5 |

2, 3, 4, 6 | 2, 3, 4, 7 | 2, 3, 5, 6 | 2, 3, 5, 7 | 2, 3, 6, 7 | 2, 4, 5, 6 | 2, 4, 5, 7 |

2, 4, 6, 7 | 2, 5, 6, 7 | 3, 4, 5, 6 | 3, 4, 5, 7 | 3, 4, 6, 7 | 3, 5, 6, 7 | 4, 5, 6, 7 |

If two white and two black balls are chosen, then their numbers form one of the following \(\frac{3\cdot 2}{1\cdot2} \times \frac{4\cdot3}{1\cdot2} = 18\) sets:

1, 2, 4, 5 | 1, 2, 4, 6 | 1, 2, 4, 7 | 1, 2, 5, 6 | 1, 2, 5, 7 | 1, 2, 6, 7 |

1, 3, 4, 5 | 1, 3, 4, 6 | 1, 3, 4, 7 | 1, 3, 5, 6 | 1, 3, 5, 7 | 1, 3, 6, 7 |

2, 3, 4, 5 | 2, 3, 4, 6 | 2, 3, 4, 7 | 2, 3, 5, 6 | 2, 3, 5, 7 | 2, 3, 6, 7 |

In this way, we have 35 equally likely cases, 18 of which favor the considered event; consequently, the desired probability that there will be two of each color among the four chosen balls is \(\frac{18}{35}\).^{5}

*Note that Markov did not use “\(!\)” for factorial or any combinatorial notation such as \(\binom{n}{k}\) or \(\,_n C_k\), although such symbols had been around since the early 19th century. There are many other places in the text where Markov's writing could be simplified if even more modern terminology, such as “sample space,” “normal distribution,” and “variance” had been invented.*

*Markov provided a second solution to this problem. It is rather convoluted and not much is gained by including it here. It can be found (together with another numerical example) in the complete translation (pp. 78*–*80).*

[5] In other words, \( \frac{\binom{3}{2}\binom{4}{2}}{\binom{7}{4}} = \frac{18}{35} \).