题目
某种产品的商标为"MAXAM",其中有2个字母脱落,有人捡起随意放-|||-回,求放回后仍为"MAX AM"的概率.
题目解答
答案
解析
步骤 1：定义事件
定义事件 $H_1, H_2, H_3, H_4, H_5$ 分别表示脱落的字母组合为 "MM", "AA", "MA", "XA", "XM"。定义事件 $G$ 表示放回后仍为 "MAX AM"。
步骤 2：计算各事件的概率
- $P(H_1) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$
- $P(H_2) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$
- $P(H_3) = \frac{\binom{2}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{4}{10}$
- $P(H_4) = \frac{\binom{1}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{2}{10}$
- $P(H_5) = \frac{\binom{1}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{2}{10}$
步骤 3：计算条件概率
- $P(G|H_1) = P(G|H_2) = 1$
- $P(G|H_3) = P(G|H_4) = P(G|H_5) = \frac{1}{2}$
步骤 4：应用全概率公式
$P(G) = \sum_{i=1}^{5} P(H_i)P(G|H_i) = \frac{1}{10} \times 1 + \frac{1}{10} \times 1 + \frac{4}{10} \times \frac{1}{2} + \frac{2}{10} \times \frac{1}{2} + \frac{2}{10} \times \frac{1}{2} = \frac{2}{5}$
定义事件 $H_1, H_2, H_3, H_4, H_5$ 分别表示脱落的字母组合为 "MM", "AA", "MA", "XA", "XM"。定义事件 $G$ 表示放回后仍为 "MAX AM"。
步骤 2：计算各事件的概率
- $P(H_1) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$
- $P(H_2) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$
- $P(H_3) = \frac{\binom{2}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{4}{10}$
- $P(H_4) = \frac{\binom{1}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{2}{10}$
- $P(H_5) = \frac{\binom{1}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{2}{10}$
步骤 3：计算条件概率
- $P(G|H_1) = P(G|H_2) = 1$
- $P(G|H_3) = P(G|H_4) = P(G|H_5) = \frac{1}{2}$
步骤 4：应用全概率公式
$P(G) = \sum_{i=1}^{5} P(H_i)P(G|H_i) = \frac{1}{10} \times 1 + \frac{1}{10} \times 1 + \frac{4}{10} \times \frac{1}{2} + \frac{2}{10} \times \frac{1}{2} + \frac{2}{10} \times \frac{1}{2} = \frac{2}{5}$