7. Seedlings of two varieties of black currant, 6 of the Selyanchenskaya variety and 8 of the Vologda variety, are prepared for planting in a garden plot and are accidentally mixed. What is the probability that the first 3 currant seedlings planted will be of the Selyanchenskaya variety?
S o l u t i o n. Let's denote the event: $A$ - the first 3 currant seedlings planted are of the Selchenkaya variety.

We will find the probability of event $A$ by applying formula (1). The numbers $m$ and $n$, which are part of this formula, will be obtained using the formulas of combination theory.

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-012.jpg?height=221&width=1052&top_left_y=1192&top_left_x=153)

Fig. 4

There are 14 elements - 14 currant seedlings. These elements are represented on Fig. 4 by the symbols $\otimes$ and $\Theta$ and are numbered from 1 to 14. On Fig. 4, the currant seedlings of the Selchenkaya variety are marked with numbers from 1 to 6, and those of the Vologda variety are marked with numbers from 7 to 14.

According to the condition, each combination of 14 elements includes 3 elements, different combinations differ from each other by at least one element, and the order of the elements does not matter. Possible combinations are: 1-3-6; 1-7-8; 3-4-6; 11-13-14, etc. Thus, the total number of equally possible outcomes of the experiment is equal to the number of combinations of 14 elements taken 3 at a time, i.e., $n=C_{14}^{3}$. Using formula (4), we find: $C_{14}^{3}=\frac{14!}{3!\cdot 11!}=364$.

Favorable to event $A$ will be combinations of 6 elements (currant seedlings of the Selchenkaya variety), each of which includes 3 elements, different combinations differ from each other by at least one element, and the order of the elements does not matter. Favorable to event $A$ will be combinations: $1-2-6 ; 4-3-2 ; 1-5-3$ and so on. Thus, the number of outcomes of the experiment favorable to event $A$ is equal to the number of combinations of 6 elements taken 3 at a time. Using formula (4), we find $m=C_{6}^{3}=\frac{6!}{3!\cdot 3!}=20$.

The sought probability of event $A$ is

$$
P(A)=\frac{m}{n}=\frac{C_{6}^{3}}{C_{14}^{3}}=\frac{20}{364}=\frac{5}{91} \approx 0.0549
$$