概率公式练习题 - 掌握加法公式与乘法公式的综合应用
以下是3道精选练习题，涵盖概率公式的各种应用场景和计算技巧。
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) \( P(A' \cup B) \)
解答过程：
a) \( P(A \cap B) = 0.4 + 0.5 - 0.6 = 0.3 \)
b) \( P(A') = 1 - 0.4 = 0.6 \)
c) \( P(B') = 0.5 \)，\( P(A \cap B') = 0.4 - 0.3 = 0.1 \)，故 \( P(A \cup B') = 0.4 + 0.5 - 0.1 = 0.8 \)
d) \( P(A' \cap B) = 0.5 - 0.3 = 0.2 \)，故 \( P(A' \cup B) = 0.6 + 0.5 - 0.2 = 0.9 \)
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.
解答过程：
设 \( F \) = 有冰箱，\( D \) = 有洗碗机。
\( P(F \cap D) = P(F) + P(D) - P(F \cup D) = 0.7 + 0.2 - 0.8 = 0.1 \)
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.
You must show all your working.
解答过程：
a) \( P(J \cap C) = 0.6 + 0.7 - 0.8 = 0.5 \)
b) \( P(C') = 0.3 \)，\( P(J \cap C') = 0.6 - 0.5 = 0.1 \)，故 \( P(J|C') = \frac{0.1}{0.3} = \frac{1}{3} \approx 0.333 \)
c) \( P(C|J) = \frac{0.5}{0.6} = \frac{5}{6} \approx 0.833 \)
d) \( P(J) \times P(C) = 0.6 \times 0.7 = 0.42 \neq 0.5 = P(J \cap C) \)，故 \( J \) and \( C \) are **not independent**.
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
解答过程：
a) \( P(A \cap B) = P(A) \times P(B|A) = 0.3 \times 0.5 = 0.15 \)
b) \( P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.15 = 0.55 \)
c) \( P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.3} = 0.5 \)
d) \( P(A) \times P(B) = 0.3 \times 0.4 = 0.12 \neq 0.15 = P(A \cap B) \)，故不独立。
假设某种疾病的发病率为0.01，已知检测阳性的概率为0.95（真阳性），假阳性率为0.05。求：
a) 检测阳性时实际患病的概率
b) 检测阴性时实际健康的概率
解答过程：
设 \( D \) = 患病，\( T \) = 检测阳性。
\( P(D) = 0.01 \)，\( P(T|D) = 0.95 \)，\( P(T|D') = 0.05 \)，\( P(D') = 0.99 \)
a) \( P(D|T) = \frac{P(T|D) \times P(D)}{P(T)} = \frac{0.95 \times 0.01}{0.95 \times 0.01 + 0.05 \times 0.99} = \frac{0.0095}{0.0095 + 0.0495} = \frac{0.0095}{0.059} \approx 0.161 \)
b) \( P(D'|T') = \frac{P(T'|D') \times P(D')}{P(T')} = \frac{0.95 \times 0.99}{0.95 \times 0.99 + 0.05 \times 0.01} = \frac{0.9405}{0.9405 + 0.0005} \approx 0.9995 \)