（12 分）一个不透明的袋子中装有 5 个小球，其中有 3 个红球， 2 个白球，这些球除颜色外完全相同.

(1) 记事件 $A$ 为“一次摸出 2 个球, 摸出的球为一个红球,一个白球”, 求 $P(A)$ ，

(2) 记事件 $B$ 为“第一次摸出一个球, 记下颜色后将它放回袋中, 再次摸出一个球,两次摸出的球为不同颜色的球”, 记事件 $C$ 为“第一次摸出一个球, 不放回袋中, 再次摸出一个球, 两次摸出的球为不同颜色的球”, 求证: $P(C)-P(B)=\frac{1}{5} P(A)$.
(1) 记这 3 个红球为 $a_{1}, a_{2}, a_{3}, 2$ 个白球记为 $b_{1}, b_{2}$,

则从袋中一次摸出 2 个球的所有基本事件为 $\left(a_{1}, a_{2}\right),\left(a_{1}, a_{3}\right),\left(a_{1}, b_{1}\right),\left(a_{1}, b_{2}\right),\left(a_{2}, a_{3}\right),\left(a_{2}, b_{1}\right)$,

$\left(a_{2}, b_{2}\right),\left(a_{3}, b_{1}\right),\left(a_{3}, b_{2}\right),\left(b_{1}, b_{2}\right)$ 共 10 个,

其中满足事件 $A$ 的基本事件有 6 个,

所以 $P(A)=\frac{6}{10}=\frac{3}{5}$.

（2）从袋中第一次摸出一个球，记下颜色后将它放回袋中，再次摸出一个球的所有基本事件为 $\left(a_{1}, a_{1}\right)$, $\left(a_{1}, a_{2}\right),\left(a_{1}, a_{3}\right),\left(a_{1}, b_{1}\right),\left(a_{1}, b_{2}\right),\left(a_{2}, a_{1}\right),\left(a_{2}, a_{2}\right),\left(a_{2}, a_{3}\right),\left(a_{2}, b_{1}\right),\left(a_{2}, b_{2}\right),\left(a_{3}, a_{1}\right)$,
$\left(a_{3}, a_{2}\right),\left(a_{3}, a_{3}\right),\left(a_{3}, b_{1}\right),\left(a_{3}, b_{2}\right),\left(b_{1}, a_{1}\right),\left(b_{1}, a_{2}\right),\left(b_{1}, a_{3}\right),\left(b_{1}, b_{1}\right),\left(b_{1}, b_{2}\right),\left(b_{2}, a_{1}\right)$,

$\left(b_{2}, a_{2}\right),\left(b_{2}, a_{3}\right),\left(b_{2}, b_{1}\right),\left(b_{2}, b_{2}\right)$ 共 25 个,

满足事件 $B$ 的基本事件有 12 个,

所以 $P(B)=\frac{12}{25}$.

从袋中第一次摸出一个球, 不放回袋中, 再次摸出一个球的所有基本事件为 $\left(a_{1}, a_{2}\right),\left(a_{1}, a_{3}\right),\left(a_{1}, b_{1}\right)$,

$\left(a_{1}, b_{2}\right),\left(a_{2}, a_{1}\right),\left(a_{2}, a_{3}\right),\left(a_{2}, b_{1}\right),\left(a_{2}, b_{2}\right),\left(a_{3}, a_{1}\right),\left(a_{3}, a_{2}\right),\left(a_{3}, b_{1}\right),\left(a_{3}, b_{2}\right),\left(b_{1}, a_{1}\right)$,

$\left(b_{1}, a_{2}\right),\left(b_{1}, a_{3}\right),\left(b_{1}, b_{2}\right),\left(b_{2}, a_{1}\right),\left(b_{2}, a_{2}\right),\left(b_{2}, a_{3}\right),\left(b_{2}, b_{1}\right)$ 共 20 个,

满足事件 $C$ 的基本事件有 12 个,

所以 $P(C)=\frac{12}{20}=\frac{3}{5}$.

因此: $P(C)-P(B)=\frac{3}{5}-\frac{12}{25}=\frac{3}{25}$,

又 $P(A)=\frac{3}{5}$, 所以 $P(C)-P(B)=\frac{1}{5} P(A)$.