1. A cage contains 10 cats, 7 of which are white and 3 are black. The cage door is opened slightly, allowing only one cat to escape at a time. The cats rush to get out. If all 10 cats eventually escape, let $X$ represent the number of segments into which the 7 white cats are divided by the 3 black cats. For example, in the order of escape, $X=3$, because the 7 white cats are divided into 3 segments. Try to find the distribution of $X$.
(Su Chun, provided)
1. 3 black cats can divide 7 white cats into at most 4 segments, so the set of values for $X$ is $\{1,2,3,4\}$. Considering the position allocation, the total number of ways $|\Omega|=\mathrm{C}_{10}^{3}=120$ (choosing 3 positions out of 10 for the black cats). For $k \in\{1,2,3,4\}$, when $X=k$, the 7 white cats are divided into $k$ segments by the 3 black cats. Which $k$ segments? There are $\mathrm{C}_{4}^{k}$ different possibilities. The next question is: how many white cats are in each segment? At this point, we consider the 7 white cats as 7 white balls arranged in a row, and we need to insert $k-1$ dividers between them, and these dividers must be placed in different positions, so there are $\mathrm{C}_{6}^{k-1}$ different ways to do this. Thus, we have
$$
\operatorname{Card}(X=k)=\mathrm{C}_{4}^{k} \cdot \mathrm{C}_{6}^{k-1} .
$$

Therefore,
$$
P(X=k)=\frac{\mathrm{C}_{4}^{k} \cdot \mathrm{C}_{6}^{k-1}}{\mathrm{C}_{10}^{3}}, \quad k=1,2,3,4 .
$$

Substituting specific values, we calculate the probabilities and present them in a table (see Table 1).
Table 1
\begin{tabular}{c|c|c|c|c}
\hline$X$ & 1 & 2 & 3 & 4 \\
\hline$P$ & $\frac{1}{30}$ & $\frac{9}{30}$ & $\frac{15}{30}$ & $\frac{5}{30}$ \\
\hline
\end{tabular}

Note: Our calculation is correct, in fact
$$
\sum_{k=1}^{4} \mathrm{C}_{4}^{k} \cdot \mathrm{C}_{6}^{k-1}=\sum_{k=1}^{4} \mathrm{C}_{4}^{4-k} \cdot \mathrm{C}_{6}^{k-1}=\mathrm{C}_{10}^{3} .
$$