13. Line-up. (From 8th grade, 3 points) Twenty kindergarten children, including 11 girls and 9 boys, stand in a line-up at random. Find the probability that there will be no more than five girls between the first and the last boy.
Solution. Let's number the places in the row from left to right. The condition of the problem is satisfied by rows in which all boys are concentrated in the first 14 places. There are exactly $C_{14}^{9}$ such combinations, and thus such rows.

In addition to these rows, the condition is satisfied by all rows where the last boy stands in the 15th place, and the other 8 boys are somehow distributed among the places from 2nd to 14th (a total of 13 places). There are $C_{13}^{8}$ such rows. The same number of rows exist where the last boy is in the 16th place, and the other 8 occupy some of the 13 places from 3rd to 15th. The same number of combinations exist when the last boy stands in the 17th, 18th, 19th, and 20th places. Thus, in total, there are $6 C_{13}^{8}$ different rows that satisfy the condition of the problem, where the last boy stands in a place numbered 15 or more.

![](https://cdn.mathpix.com/cropped/2024_05_06_11f8b78b86004d31b3e4g-10.jpg?height=500&width=1222&top_left_y=521&top_left_x=494)

In total, 9 boys can be arranged in the row in $C_{20}^{9}$ ways, of which $C_{14}^{9}+6 C_{13}^{8}$ ways satisfy the condition of the problem. Assuming all combinations are equally likely (as the children lined up randomly), we find the desired probability:

$$
\frac{C_{14}^{9}+6 C_{13}^{8}}{C_{20}^{9}} \approx 0.058
$$

Answer: approximately 0.058.