掌握加法公式与乘法公式的综合应用 - 知识点总结与练习题
重要概念：对于任意两个事件A和B，并集概率与交集概率的关系
重要概念：条件概率与交集概率的关系
重要概念：通过比较交集概率与独立情况下的乘积来判断事件独立性
公式重排：
| 情况 | 条件 | 加法公式简化 | 乘法公式简化 |
|---|---|---|---|
| 互斥事件 | \( P(A \cap B) = 0 \) | \( P(A \cup B) = P(A) + P(B) \) | 无简化 |
| 独立事件 | \( P(A \cap B) = P(A)P(B) \) | 无简化 | \( P(A \cap B) = P(A)P(B) \) |
题目：已知 \( P(A) = 0.4 \), \( P(B) = 0.5 \) and \( P(A \cup B) = 0.6 \)，求 \( P(A \cap B) \)
使用加法公式重排：
题目：已知 \( P(A) = 0.6 \)，\( P(C) = 0.7 \) and \( P(A \cup C) = 0.8 \)，判断A和C是否独立
步骤1： 计算交集概率
\( P(A \cap C) = P(A) + P(C) - P(A \cup C) = 0.6 + 0.7 - 0.8 = 0.5 \)
步骤2： 计算独立情况下的乘积
\( P(A) imes P(C) = 0.6 imes 0.7 = 0.42 \)
步骤3： 比较
由于 \( 0.5 eq 0.42 \)，所以A和C不独立
A and B are two events where \( P(A) = 0.4 \), \( P(B) = 0.5 \) and \( P(A \cup B) = 0.6 \)
Find:
a) \( P(A \cap B) \)
b) \( P(A') \)
c) \( P(A \cup B') \)
d) \( P(A' \cup B) \)
答题区域：
A survey of a large number of households in Istanbul was carried out. The survey showed that 70% have a freezer, 20% have a dishwasher and 80% have either a dishwasher or a freezer or both appliances. Find the probability that a randomly chosen household in Istanbul has both appliances.
答题区域：
假设有两个事件A和B，已知\( P(A) = 0.3 \)，\( P(B) = 0.4 \)，\( P(A|B) = 0.5 \)。求：
a) \( P(A \cap B) \)
b) \( P(A \cup B) \)
c) \( P(B|A) \)
d) 判断A和B是否独立
答题区域：
José and Cristiana play darts on the same team. The events \( J \) and \( C \) are defined as follows:
\( J \) is the event that José wins his match.
\( C \) is the event that Cristiana wins her match.
\( P(J) = 0.6 \)，\( P(C) = 0.7 \) and \( P(J \cup C) = 0.8 \)
Find the probability that:
a) both José and Cristiana win their matches
b) José wins his match given that Cristiana loses hers
c) Cristiana wins her match given that José wins his.
d) Determine whether or not the events \( J \) and \( C \) are independent.
答题区域：
解答过程：
a) 使用加法公式：\( P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.6 = 0.3 \)
b) 互补概率：\( P(A') = 1 - P(A) = 1 - 0.4 = 0.6 \)
c) \( P(B') = 1 - P(B) = 0.5 \)，\( P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.3 = 0.1 \)
故 \( P(A \cup B') = P(A) + P(B') - P(A \cap B') = 0.4 + 0.5 - 0.1 = 0.8 \)
d) \( P(A' \cap B) = P(B) - P(A \cap B) = 0.5 - 0.3 = 0.2 \)
故 \( P(A' \cup B) = P(A') + P(B) - P(A' \cap B) = 0.6 + 0.5 - 0.2 = 0.9 \)
解答过程：
设 \( F \) = 有冰箱，\( D \) = 有洗碗机。
已知 \( P(F) = 0.7 \)，\( P(D) = 0.2 \)，\( P(F \cup D) = 0.8 \)
使用加法公式：\( P(F \cap D) = P(F) + P(D) - P(F \cup D) = 0.7 + 0.2 - 0.8 = 0.1 \)
解答过程：
a) 使用乘法公式：\( P(A \cap B) = P(B) imes P(A|B) = 0.4 imes 0.5 = 0.2 \)
b) 使用加法公式：\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.2 = 0.5 \)
c) 使用条件概率公式：\( P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.3} = \frac{2}{3} \approx 0.667 \)
d) 独立性判断：\( P(A) imes P(B) = 0.3 imes 0.4 = 0.12 eq 0.2 = P(A \cap B) \)，故不独立。
解答过程：
a) 使用加法公式：\( P(J \cap C) = P(J) + P(C) - P(J \cup C) = 0.6 + 0.7 - 0.8 = 0.5 \)
b) \( P(C') = 1 - 0.7 = 0.3 \)，\( P(J \cap C') = P(J) - P(J \cap C) = 0.6 - 0.5 = 0.1 \)
故 \( P(J|C') = \frac{P(J \cap C')}{P(C')} = \frac{0.1}{0.3} = \frac{1}{3} \approx 0.333 \)
c) \( P(C|J) = \frac{P(J \cap C)}{P(J)} = \frac{0.5}{0.6} = \frac{5}{6} \approx 0.833 \)
d) 独立性判断：\( P(J) imes P(C) = 0.6 imes 0.7 = 0.42 eq 0.5 = P(J \cap C) \)
故 \( J \) and \( C \) are not independent.