随机变量 $X$ 服从二项分布 $\text{B}(n, p)$，其中 $0 < p < 1$，且 $n$ 为整数。证明
$\frac{\text{P}(X = r)}{\text{P}(X = r - 1)} = \left( \frac{n - r + 1}{r} \right) \left( \frac{p}{1 - p} \right).$

1. Equate the given ratio to 1 for the existence of two adjacent equal probabilities

For $X$ to have two values for its mode, there must be two adjacent integer values where the probabilities are equal and at their maximum. Let these two values be $r$ and $r - 1$.
$\frac{\text{P}(X = r)}{\text{P}(X = r - 1)} = 1$

2. Solve for $r$ in terms of $n$ and $p$

$\left( \frac{n - r + 1}{r} \right) \left( \frac{p}{1 - p} \right) = 1$
$p(n - r + 1) = r(1 - p)$
$pn - pr + p = r - pr$
$r = pn + p = (n + 1)p$

3. Deduce the condition for $n$ and $p$, and state the two modes

Since $r$ represents a possible value of $X$, it must be an integer.
Therefore, the condition is that $(n + 1)p$ is an integer.
The two values for the mode are $r$ and $r-1$, which correspond to $(n + 1)p$ and $(n + 1)p - 1$.