9. Fly on the Grid. A fly crawls from the origin. In doing so, the fly moves only along the lines of the integer grid to the right or up (monotonic walk). At each node of the grid, the fly randomly chooses the direction of further movement: up or to the right. Find the probability that at some point:

a) the fly will be at the point $(8 ; 10)$;

b) the fly will be at the point $(8 ; 10)$, having passed through the segment connecting the points $(5 ; 6)$ and $(6 ; 6)$;

c) the fly will be at the point $(8 ; 10)$, having passed inside a circle of radius 3 centered at the point $(4 ; 5)$.
Solution. a) The fly can reach the point $(8 ; 10)$ in exactly 18 steps - no more, no less. Therefore, it is sufficient to consider the first 18 steps of the fly. The total number of possible paths consisting of 18 steps is $2^{18}$.

To reach the point $(8 ; 10)$ from these 18 steps, the fly must take exactly 8 steps to the right and exactly 10 steps up. The order in which the steps to the right and up alternate is not important. Therefore, the number of paths leading to the point $(8 ; 10)$ is $C_{18}^{8}$.

Thus, the probability of reaching this point is $\frac{C_{18}^{8}}{2^{18}} \approx 0.167$.

b) Similarly to the previous problem, we will find the number of paths leading to the point $(5 ; 6)$. This number is $C_{11}^{5}$. There is only one path from the point $(5 ; 6)$ to the point $(6 ; 6)$. Then the fly must go from the point $(6 ; 6)$ to the point $(8 ; 10)$, and here it has $C_{6}^{2}$ possible paths. Therefore, the total number of paths satisfying the condition of the problem is $C_{11}^{5} \cdot C_{6}^{2}$.

The total number of paths consisting of 18 steps is $2^{18}$. Therefore, the probability of reaching the point $(8 ; 10)$, passing through the points $(5 ; 6)$ and $(6 ; 6)$ along the way, is $\frac{C_{11}^{5} \cdot C_{6}^{2}}{2^{18}} \approx 0.026$.

c) The fly will pass inside the circle if and only if it passes through one of the five marked points (see figure). The fly can pass through only one of these points. Therefore, to find the number of paths leading to the point $(8 ; 10)$ through the circle, we need to find the number of paths passing through each of these points and sum the results:

![](https://cdn.mathpix.com/cropped/2024_05_06_c76ac3d7d759b54fc6bag-10.jpg?height=716&width=646&top_left_y=1975&top_left_x=1276)

\[
\begin{aligned}
& C_{9}^{2} C_{9}^{6} + C_{9}^{3} C_{9}^{5} + C_{9}^{4} C_{9}^{4} + C_{9}^{5} C_{9}^{3} + C_{9}^{6} C_{9}^{2} = \\
& = 2 C_{9}^{2} C_{9}^{6} + 2 C_{9}^{3} C_{9}^{5} + C_{9}^{4} C_{9}^{4}
\end{aligned}
\]

The desired probability is

\[
\frac{2 C_{9}^{2} C_{9}^{6} + 2 C_{9}^{3} C_{9}^{5} + C_{9}^{4} C_{9}^{4}}{2^{18}} = \frac{43092}{2^{18}} = \frac{10773}{2^{16}} \approx 0.164
\]

Answer: a) $\frac{C_{18}^{8}}{2^{18}} \approx 0.167$;

b) $\frac{C_{11}^{5} \cdot C_{6}^{2}}{2^{18}} \approx 0.026$;

c) $\frac{2 C_{9}^{2} C_{9}^{6} + 2 C_{9}^{3} C_{9}^{5} + C_{9}^{4} C_{9}^{4}}{2^{18}} \approx 0.164$.