【Example 1】(Ticket Selling Problem) Each football ticket costs 50 yuan, and the ticket office has no change. There are $m$ people each holding a 50-yuan note, and another $n$ people each holding a 100-yuan note $(m \geqslant n)$ queuing to buy tickets, with each person buying only one ticket. If no one has to wait for change, it is called a smooth ticket purchase. Find the probability of a smooth ticket purchase.
Let's discuss the methods for finding $|\Omega|$ and $|E|$. To ensure a smooth ticket purchase, the number of people holding 50 yuan must not be less than the number of people holding 100 yuan up to any point in the queue.

Let $y_{k}$ represent the difference between the number of people holding 50 yuan and the number of people holding 100 yuan among the first $k$ people. Each queue arrangement corresponds to a sequence $\left(y_{0}, y_{1}, \cdots, y_{m+n}\right)$, where $\Omega$ consists of all such sequences. Clearly, for each such sequence, we have $y_{0}=0, y_{m+n}=m-n$, and for any $k \in\{1,2, \cdots, m+n\}$,
$$
y_{k}-y_{k-1}=\left\{\begin{aligned}
1, & \text { if the } k \text {th person holds } 50 \text { yuan } \\
-1, & \text { if the } k \text {th person holds } 100 \text { yuan }
\end{aligned}\right.
$$

To study these sequences, we establish a Cartesian coordinate system. For any sequence $\left(y_{0}, y_{1}, \cdots, y_{m+n}\right)$, we mark points $\left(0, y_{0}\right),\left(1, y_{1}\right),\left(2, y_{2}\right), \cdots,\left(m+n, y_{m+n}\right)$ in the coordinate system and connect these points with line segments, resulting in a broken line starting from $\left(0, y_{0}\right)=(0,0)$ and ending at $(m+n, y_{m+n}) = (m+n, m-n)$ (refer to Figure 17). We call each line segment connecting points $\left(k-1, y_{k-1}\right)$ and $\left(k, y_{k}\right)$ a segment. Each such broken line has $m+n$ segments, each with a slope of -1 or 1, and each segment has a length of $\sqrt{2}$. We call each such broken line a path.
On each such path, there are $m$ segments with a slope of 1 and $n$ segments with a slope of -1. The number of different paths is given by the permutations of distinct elements, so the total number of different such paths is $C_{m+n}^{n}$. Since $|\Omega|$ is the total number of different such paths, we have
$$
|\Omega|=C_{m+n}^{m}
$$

To ensure a smooth ticket purchase, it must be that $y_{1}=1, y_{2} \geqslant 0, \cdots, y_{m+n} \geqslant 0$, so
$$
E=\left\{\left(y_{0}, y_{1}, \cdots, y_{m+n}\right): y_{1}=1, y_{2} \geqslant 0, \cdots, y_{m+n} \geqslant 0\right\} .
$$

If the sequence $\left(y_{0}, y_{1}, \cdots, y_{m+n}\right) \in E$, then the corresponding path lies above the x-axis, possibly intersecting the x-axis but not the line $y=-1$. Our task is to find the number of such paths $|E|$.

If we denote $E^{\mathrm{c}}=\Omega \backslash E$, i.e., $E^{\mathrm{c}}$ represents the set of all paths starting from $A(0,0)$ and ending at $B(m+n, m-n)$ that intersect the line $y=-1$, then $|E| = |\Omega| - \left|E^{c}\right| = C_{m+n}^{m} - \left|E^{c}\right|$. Therefore, to find $|E|$, we only need to find $\left|E^{c}\right|$.

Take any path in $E^{c}$, and let the leftmost intersection point of this path with the line $y=-1$ be $T$ (see Figure 18). Reflect the part of the path between points $A$ and $T$ about the line $y=-1$, resulting in a path starting from $A^{\prime}(0, -2)$ and ending at $B(m+n, m-n)$. Conversely, since $m \geqslant n$, any path starting from $A^{\prime}(0, -2)$ and ending at $B(m+n, m-n)$ must intersect the line $y=-1$. If we reflect the part of such a path between point $A^{\prime}$ and its leftmost intersection with $y=-1$ about the line $y=-1$, we obtain a path starting from $A(0,0)$ and ending at $B(m+n, m-n)$ that intersects the line $y=-1$. Therefore, there is a one-to-one correspondence between these two types of paths, and their numbers are equal.

Any path starting from $A^{\prime}(0, -2)$ and ending at $B(m+n, m-n)$ is a broken line with $m+n$ segments. Let the number of segments with a slope of 1 be $u$ and the number of segments with a slope of -1 be $v$. Then we have
$$
\left\{\begin{array}{l}
u+v=m+n \\
u-v=m-n+2
\end{array}\right.
$$

So $u=m+1$. Therefore, $\left|E^{c}\right|=C_{m+n}^{m+1}$,
$$
\begin{aligned}
|E| & =|\Omega|-\left|E^{c}\right|=C_{m+n}^{n}-C_{m+n}^{m+1} \\
& =\frac{(m+n)!}{m!n!}-\frac{(m+n)!}{(m+1)!(n-1)!} \\
& =\frac{m-n+1}{m+1} C_{m+n}^{n}
\end{aligned}
$$

Substituting back into (1), we get
$$
P(E)=\frac{|E|}{|\Omega|}=\frac{m-n+1}{m+1}
$$