Example 10. Two shooters make one shot each at one target. The probability of hitting the target for the first shooter with one shot $p_{1}=0.5$, for the second - $p_{2}=0.4$. The discrete random variable $X$ is the number of hits on the target. Find the distribution function of this random variable. Find the probability of the event $X \geq 1$.
Solution. First, let's find the distribution law of the given discrete random variable $X$. This variable can take three values: $x_{1}=0, x_{2}=1, x_{3}=2$.

We introduce the notation: event $A_{1}$ - "the first shooter hits", event $A_{2}$ - "the second shooter hits", then $\bar{A}_{1}$ and $\bar{A}_{2}$ are their respective misses. From the problem statement, it follows that

$$
\begin{aligned}
& q_{1}=P\left(\overline{A_{1}}\right)=1-p_{1}=0.5 \\
& q_{2}=P\left(\bar{A}_{2}\right)=1-p_{2}=0.6
\end{aligned}
$$

The value $x_{1}=0$ corresponds to the case when both shooters miss - the event $A=\bar{A}_{1} \bar{A}_{2}$, where $\bar{A}_{1}$ and $\bar{A}_{2}$ are independent events since $A_{1}$ and $A_{2}$ are independent.

By the multiplication theorem, we get

$$
P(A)=P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right)=0.5 \cdot 0.6=0.3
$$

The value $x_{2}=1$ corresponds to the case when the number of hits is one: the first shooter hits and the second misses, or the second hits and the first misses. This means that the event

$$
B=A_{1} \bar{A}_{2}+A_{2} \bar{A}_{1}
$$

occurred, where $A_{1} \bar{A}_{2}$ and $A_{2} \bar{A}_{1}$ are mutually exclusive events, and $A_{1}$ and $\bar{A}_{2}$, $A_{2}$ and $\bar{A}_{1}$ are independent events, respectively. Based on the addition and multiplication theorems, we find:

$$
\begin{aligned}
& P(B)=P\left(A_{1} \bar{A}_{2}\right)+P\left(A_{2} \bar{A}_{1}\right)=P\left(A_{1}\right) P\left(\overline{A_{2}}\right)+P\left(A_{2}\right) P\left(\overline{A_{1}}\right)= \\
& =0.5 \cdot 0.6+0.4 \cdot 0.5=0.3+0.2=0.5
\end{aligned}
$$

The value $x_{3}=2$ corresponds to the case when both shooters hit: the event $C=A_{1} A_{2}$ occurred. Therefore,

$$
P(C)=P\left(A_{1} A_{2}\right)=P\left(A_{1}\right) P\left(A_{2}\right)=0.5 \cdot 0.4=0.2
$$

Thus, the distribution law of the given random variable can be represented by the table

| $X$ | 0 | 1 | 2 |
| :---: | :---: | :---: | :---: |
| $P$ | 0.3 | 0.5 | 0.2 |

We will construct the distribution function of this random variable. Based on formula (2.2.11), we get:

1. For $x \leq 0 \quad F(x)=\sum_{x_{k}2$

$$
\begin{aligned}
& F(x)=P(X=0)+P(X=1)+P(X=2)= \\
& \quad=0.3+0.5+0.2=1
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-102.jpg?height=429&width=896&top_left_y=599&top_left_x=106)

Figure 2.8

The graph of the distribution function is shown in Figure 2.8.

We find the probability of the event $X \geq 1$. This event is equal to the sum of two events $X=1, X=2$. Therefore,

$$
P(X \geq 1)=P(X=1)+P(X=2)=0.5+0.2=0.7
$$

Example 11. A fair coin is tossed three times. Find the distribution function of the random variable $X$, which is equal to the number of heads that appear.

Solution. The given discrete random variable can take four values: $x_{1}=0, x_{2}=1, x_{3}=2, x_{4}=3$.

Let $A_{1}, A_{2}, A_{3}$ be the events "a head appears" on the first, second, and third tosses, respectively. In this experiment, there are 8 elementary outcomes: $\bar{A}_{1} \bar{A}_{2} \bar{A}_{3}, A_{1} \bar{A}_{2} \bar{A}_{3}, \bar{A}_{1} A_{2} \bar{A}_{3}, \bar{A}_{1} \bar{A}_{2} A_{3}, \bar{A}_{1} A_{2} A_{3}$, $A_{1} \bar{A}_{2} A_{3}, A_{1} A_{2} \bar{A}_{3}, A_{1} A_{2} A_{3}$

Considering the independence of events $A_{1}, A_{2}, A_{3}$ and their complementary events $\overline{A_{1}}, \bar{A}_{2}, \bar{A}_{3}$, and that the probability of a head appearing on a single toss is $1 / 2$, we find the corresponding probabilities:

$$
\begin{aligned}
& P(X=0)=P\left(\bar{A}_{1} \bar