Example. Two dice are thrown. Find the probability that the number of points on one die is twice the number on the other.
Solution. When the number of equally possible elementary outcomes is finite, the probability of an event is equal to the ratio of the number of elementary outcomes favorable to this event to the total number of elementary outcomes.

1. Define the space of elementary outcomes.

Number the dice. Then an elementary outcome of the random experiment is an ordered pair of numbers $\left(n_{1}, n_{2}\right)$, where $n_{1}$ is the number of points that fell on the first die and $n_{2}$ on the second.

All elementary outcomes can be conveniently illustrated in the form of a table

| $(1,1)$ | $(1,2)$ | $(1,3)$ | $(1,4)$ | $(1,5)$ | $(1,6)$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $(2,1)$ | $(2,2)$ | $(2,3)$ | $(2,4)$ | $(2,5)$ | $(2,6)$ |
| $(3,1)$ | $(3,2)$ | $(3,3)$ | $(3,4)$ | $(3,5)$ | $(3,6)$ |
| $(4,1)$ | $(4,2)$ | $(4,3)$ | $(4,4)$ | $(4,5)$ | $(4,6)$ |
| $(5,1)$ | $(5,2)$ | $(5,3)$ | $(5,4)$ | $(5,5)$ | $(5,6)$ |
| $(6,1)$ | $(6,2)$ | $(6,3)$ | $(6,4)$ | $(6,5)$ | $(6,6)$ |

2. Check the equiprobability of elementary outcomes.

Since the dice are symmetrical, made of homogeneous material, and thrown properly, all elementary outcomes are equiprobable.

3. Select those elementary outcomes that lead to the occurrence of event $A$ (favorable outcomes).

For the event $A$ of interest, the favorable outcomes are those $\left(n_{1}, n_{2}\right)$ for which $n_{1}=2 n_{2}$ or $2 n_{1}=n_{2}$. Clearly, these are the outcomes $(1,2),(2,4),(3,6),(2,1),(4,2)$, and $(6,3)$.

4. Count the total number of elementary outcomes $N$ (the number of elements in the space of elementary outcomes). Clearly, $N=36$.
5. Count the number of elementary outcomes favorable to event $A$. Clearly, $N_{A}=6$.
6. According to the classical definition of probability

$$
\mathrm{P}(A)=\frac{N_{A}}{N}=\frac{6}{36}
$$

Answer. $\mathrm{P}(A)=6 / 36$.

Remark. If the dice are indistinguishable, it seems quite natural to consider an unordered set of two numbers that fell as an elementary outcome, writing them, for example, in ascending order. However, in this case, the elementary outcomes will not be equiprobable (as experience shows, the outcome (1,2), for example, occurs about twice as often as the outcome $(1,1))$.

Conditions of the problem. Two dice are thrown. Find the probability of the specified event.

1. The sum of the number of points is 7.
2. The sum of the number of points is greater than 3.
3. The sum of the number of points is greater than 4 but less than 7.
4. The absolute difference of the number of points is 2.
5. The absolute difference of the number of points is greater than 1.
6. The product of the number of points is not more than 10.
7. The larger number of points is greater than 4.
8. The smaller number of points is greater than 4.
9. The number of points on at least one die is even.
10. The number of points on both dice is odd.
Answers. 1. $1 / 6$.
11. $11 / 12$. $\quad 3.1 / 4$.
12. $2 / 9$.
13. $5 / 9$.
14. $19 / 36$. 7. $4 / 9.8 .1 / 9.9 .3 / 4.10 .1 / 4$.

## 6.2. Hypergeometric Formula

Problem Statement. There is a set $D$ consisting of $n$ elements. The set $D$ is divided into two subsets $A$ and $B$ according to some criterion, where subset $A$ has $n_{A}$ elements, and subset $B$ has $n_{B}$ elements $\left(n_{A}+n_{B}=n\right)$. From $D$, $m(m \leqslant n)$ elements are randomly selected. Find the probability that among the selected $m$ elements, $m_{1}$, or $m_{2}, \ldots$, or $m_{k}\left(m_{i} \leqslant n_{1}\right)$ will belong to subset $A$.

Plan of Solution.

When the number of equally possible elementary outcomes is finite, the probability of an event is equal to the ratio of the number of elementary outcomes favorable to this event to the total number of elementary outcomes.

We will solve the problem for $k=1$, and then for $k>1$.
I. $k=1$.

1. Define the space of elementary outcomes.

An elementary outcome is an arbitrary subset of $D$ containing $m$ elements.

2. Check the equiprobability of elementary outcomes.

In this case, the equiprobability of elementary outcomes is determined by the method of selecting $m$ elements. We assume that this method guarantees the equiprobability of elementary outcomes. Note that with an unsuccessful selection method, the outcomes may not be equiprobable.

3. Select those elementary outcomes that lead to the occurrence of the event (favorable outcomes).

For this, we use the criterion by which the set $D$ is divided into two subsets $A$ and $B$. Favorable outcomes are formed by arbitrarily selecting $m_{1}$ elements from set $A$ and arbitrarily selecting $m-m_{1}$ elements from set $B$.

4. Count the total number of elementary outcomes $N$ (the number of elements in the space of elementary outcomes).

Since an elementary outcome is an arbitrary subset of $D$ containing $m$ elements, the number of elementary outcomes is equal to the number of such subsets, i.e.,

$$
N=C_{n}^{m}=\frac{n!}{m!(n-m)!}
$$

5. Count the number of elementary outcomes $N\left(m_{1}\right)$ favorable to the given event.

For this, we determine the number of elements in sets $A$ and $B$, i.e., the numbers $n_{A}$ and $n_{B}\left(n_{A}+n_{B}=n\right)$.

Favorable outcomes are formed by arbitrarily selecting $m_{1}$ elements from set $A$, consisting of $n_{A}$ elements, and arbitrarily selecting $m-m_{1}$ elements from set $B$, consisting of $n_{B}$ elements. Therefore, the number of favorable outcomes is

$$
N\left(m_{1}\right)=C_{n_{A}}^{m_{1}} C_{n_{B}}^{m-m_{1}}
$$

6. Let $P\left(m_{1}\right)$ be the probability that among the selected $m$ elements, exactly $m_{1}$ will belong to set $A$. According to the classical definition of probability

$$
P\left(m_{1}\right)=\frac{N\left(m_{1}\right)}{N}=\frac{C_{n_{A}}^{m_{1}} C_{n_{B}}^{m-m_{1}}}{C_{n}^{m}}
$$

This formula is called the hypergeometric formula.

II. $k>1$.

The number of favorable outcomes is

$$
N\left(m_{1}\right)+N\left(m_{2}\right)+\ldots+N\left(m_{k}\right)
$$

Therefore, the desired probability is

$$
\begin{array}{r}
\frac{N\left(m_{1}\right)+N\left(m_{2}\right)+\ldots+N\left(m_{k}\right)}{N}=P\left(m_{1}\right)+P\left(m_{2}\right)+\ldots+P\left(m_{k}\right)= \\
=\frac{C_{n_{A}}^{m_{1}} C_{n_{B}}^{m-m_{1}}+C_{n_{A}}^{m_{2}} C_{n_{B}}^{m-m_{2}}+\ldots+C_{n_{A}}^{m_{k}} C_{n_{B}}^{m-m_{k}}}{C_{n}^{m}}
\end{array}
$$

Remark. If $k$ is large, it is convenient to switch from the event of interest to the opposite event, taking into account that the sum of the probabilities of these two events is 1.