Problem 6. In a batch of $N$ items, there are $M(M \leqslant N)$ marked items. $n(n \leqslant N)$ items are randomly selected from the given batch. Find the probability that among these $n$ items, there will be exactly $m$ $(m \leqslant n, m \leqslant M$) marked items (event $A$).
Solution. The total number of cases in which event $A$ can occur is $n_{0}=C_{N}^{n}$.

To count the number of cases favorable to event $A$, consider a sample of $n$ items, among which exactly $m$ are marked. The total number of outcomes in which $m$ items out of $M$ can enter the sample is $C_{M}^{m}$. Each such combination of marked items can enter a sample of $n$ items, containing any combination of $(n-m)$ items from $(N-M)$ unmarked items. The number of different combinations of unmarked items in the sample is $C_{N-M}^{n-m}$. Therefore, the total number of outcomes favorable to the occurrence of event $A$ is $m_{0}=C_{M}^{m} C_{N-M}^{n-m}$. By formula (1), we get:

$$
p(A)=\frac{m_{0}}{n_{0}}=\frac{C_{M}^{m} C_{N-M}^{n-m}}{C_{N}^{n}}
$$

Problem 7. Three cards are randomly chosen from a deck of 36 cards. What is the probability that among them there will be 2 aces (event $A$)?

Solution. Three cards out of 36 can be chosen in $C_{36}^{3}$ ways, $n=C_{36}^{3}$. Two aces out of four can be chosen in $C_{4}^{2}$ ways, and this combination can be paired with any of the 32 cards that are not aces, i.e., $m=32 C_{4}^{2}$. Thus,

$$
p(A)=\frac{C_{4}^{2} \cdot 32}{C_{36}^{3}}=\frac{4 \cdot 3 \cdot 32 \cdot 1 \cdot 2 \cdot 3}{1 \cdot 2 \cdot 36 \cdot 35 \cdot 34}=\frac{16}{595} \approx 0.0269
$$

Note. From the definition, it follows that the probability of an event is a common fraction: $p(A)=\frac{m}{n}$. Sometimes it is difficult to write the exact numerical value of this fraction, for example, $p(A)=\frac{C_{20}^{5} C_{30}^{25}}{C_{50}^{30}}$. Therefore, we prefer to convert the obtained fraction to a decimal with an accuracy of three to four (sometimes more or less, depending on the context of the problem) digits after the decimal point, following the rules of rounding numbers. However, we will not write the approximate equality sign every time. For example, $p(A)=\frac{1}{7}=0.14$, $p(A)=\frac{1}{7}=0.143$, $E k=\frac{m_{4}}{\sigma^{4}}-3=\frac{79.582}{(\sqrt{4.87})^{4}}-3=0.36$.

Problem 8. On a segment $O A$ of length $L$ on the number line $O x$, a point $B(x)$ is taken at random. Find the probability that each of the segments $O B$ and $B A$ has a length greater than $\frac{L}{4}$.

Solution. Divide the segment $O A$ into four equal parts

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-133.jpg?height=92&width=402&top_left_y=1541&top_left_x=80)
(see Fig. 6.1). The requirement of the problem will be met if point $B$ falls on segment $D E$ or $C D$, the combined length of which is $\frac{L}{2}$. Therefore, by formula $5^{\circ}$, we get $p=\frac{L}{2}: L=\frac{1}{2}$.