互斥事件与独立事件练习题 - 掌握概率加法规则与乘法规则
以下是10道综合练习题，涵盖互斥事件和独立事件的定义、概率计算和独立性判断等核心内容。
Events \( A \) and \( B \) are mutually exclusive. \( P(A) = 0.2 \) and \( P(B) = 0.5 \)
a) Draw a Venn diagram to represent these two events.
b) Find \( P(A \text{ or } B) \).
c) Find \( P(\text{neither } A \text{ nor } B) \).
解答过程：
a) 维恩图
互斥事件无交集，维恩图中两个圆形不相交。
b) \( P(A \text{ or } B) \)
因互斥：\( P(A \text{ or } B) = P(A) + P(B) = 0.2 + 0.5 = 0.7 \)
c) \( P(\text{neither } A \text{ nor } B) \)
\( P(\text{neither}) = 1 - P(A \text{ or } B) = 1 - 0.7 = 0.3 \)
Two fair dice are rolled and the result on each one is recorded. Show that the events 'the sum of the scores on the dice is 4' and 'both dice land on the same number' are not mutually exclusive.
解答过程：
事件A：两个骰子点数相同
事件B：两个骰子点数和为4
证明两者不互斥：
当两个骰子都显示2时，和为4且点数相同。
这种情况同时满足A和B，故不互斥。
\( P(A) = 0.5 \) and \( P(B) = 0.3 \). Given that events \( A \) and \( B \) are independent, find \( P(A \text{ and } B) \).
解答过程：
因\( A \)、\( B \)独立，故：
\( P(A \text{ and } B) = P(A) \times P(B) = 0.5 \times 0.3 = 0.15 \)
\( P(A) = 0.15 \) and \( P(A \text{ and } B) = 0.045 \). Given that events \( A \) and \( B \) are independent, find \( P(B) \).
解答过程：
因\( A \)、\( B \)独立，故\( P(A \text{ and } B) = P(A) \times P(B) \)，即：
\( 0.045 = 0.15 \times P(B) \implies P(B) = \frac{0.045}{0.15} = 0.3 \)
The Venn diagram shows the number of children in a play group who like playing with bricks (\( B \)), action figures (\( F \)) or trains (\( T \)).
a) State, with a reason, which two types of toy are mutually exclusive.
b) Determine whether or not the events 'plays with bricks' and 'plays with action figures' are independent.
解答过程：
a) 互斥的玩具类型
假设维恩图显示某些区域为0，则这些玩具类型互斥。
需要具体维恩图数据来判断。
b) 独立性判断
计算\( P(B) \)、\( P(F) \)、\( P(B \cap F) \)，验证是否相等。
The Venn diagram shows the probabilities that a group of students like pasta (\( A \)) or pizza (\( B \)).
a) Write down the value of \( x \).
b) Determine whether the events 'likes pasta' and 'likes pizza' are independent.
解答过程：
a) 求x的值
假设样本空间概率和为1，计算缺失概率。
b) 独立性判断
计算\( P(A) \times P(B) \)与\( P(A \cap B) \)比较。
\( S \) and \( T \) are two events such that \( P(S) = 0.3 \), \( P(T) = 0.4 \) and \( P(S \text{ but not } T) = 0.18 \)
a) Show that \( S \) and \( T \) are independent.
b) Find:
i) \( P(S \text{ and } T) \)
ii) \( P(\text{neither } S \text{ nor } T) \)
解答过程：
a) 验证独立性
\( P(S \text{ but not } T) = P(S) - P(S \text{ and } T) \implies 0.18 = 0.3 - P(S \text{ and } T) \implies P(S \text{ and } T) = 0.12 \)
验证：\( P(S) \times P(T) = 0.3 \times 0.4 = 0.12 \)，相等，故独立。
b) 计算概率
i) \( P(S \text{ and } T) = 0.12 \)
ii) \( P(\text{neither}) = 1 - P(S \cup T) = 1 - (P(S) + P(T) - P(S \text{ and } T)) = 1 - (0.3 + 0.4 - 0.12) = 1 - 0.58 = 0.42 \)
\( W \) and \( X \) are two events such that \( P(W) = 0.5 \), \( P(W \text{ and not } X) = 0.25 \) and \( P(\text{neither } W \text{ nor } X) = 0.3 \). State, with a reason, whether \( W \) and \( X \) are independent events.
解答过程：
计算各概率
\( P(W) = 0.5 \)
\( P(W \text{ and not } X) = P(W) - P(W \text{ and } X) = 0.25 \implies P(W \text{ and } X) = 0.5 - 0.25 = 0.25 \)
\( P(\text{neither}) = 0.3 \implies P(W \cup X) = 1 - 0.3 = 0.7 \)
验证并集：\( P(W \cup X) = P(W) + P(X) - P(W \cap X) = 0.7 \)
\( 0.5 + P(X) - 0.25 = 0.7 \implies 0.25 + P(X) = 0.7 \implies P(X) = 0.45 \)
验证独立性
\( P(W) \times P(X) = 0.5 \times 0.45 = 0.225 \)
\( P(W \cap X) = 0.25 \neq 0.225 \)，故不独立。
The Venn diagram shows the probabilities of members of a social club taking part in charitable activities.
\( A \) represents taking part in an archery competition.
\( R \) represents taking part in a raffle.
\( F \) represents taking part in a fun run.
The probability that a member takes part in the archery competition or the raffle is 0.6.
a) Find the value of \( x \) and the value of \( y \).
b) Show that events \( R \) and \( F \) are not independent.
解答过程：
a) 求x和y
已知\( P(A \cup R) = 0.6 \)
由并集公式：\( P(A \cup R) = P(A) + P(R) - P(A \cap R) \)
需要维恩图具体数据来计算。
b) 验证独立性
计算\( P(R) \)、\( P(F) \)、\( P(R \cap F) \)，比较是否相等。
In the Venn diagram shown here, given that events \( A \) and \( B \) are independent, find the two possible values for \( p \) and \( q \).
解答过程：
假设维恩图有p和q两个未知概率。
因A和B独立，故\( P(A \cap B) = P(A) \times P(B) \)
解方程求p和q的可能值。
\( A \) and \( B \) are independent events in a sample space \( \mathcal{E} \).
Given that \( A \) and \( B \) are independent, prove that:
a) \( A \) and 'not \( B \)' are independent
b) 'not \( A \)' and 'not \( B \)' are independent.
证明过程：
a) 证明A和非B独立
要证明：\( P(A \cap B') = P(A) \times P(B') \)
左边：\( P(A \cap B') = P(A) - P(A \cap B) \)
右边：\( P(A) \times P(B') = P(A) \times (1 - P(B)) \)
因为\( P(A \cap B) = P(A) \times P(B) \)（独立），故：
\( P(A) - P(A) \times P(B) = P(A) (1 - P(B)) = P(A) \times P(B') \)，相等。
b) 证明非A和非B独立
要证明：\( P(A' \cap B') = P(A') \times P(B') \)
左边：\( P(A' \cap B') = 1 - P(A \cup B) \)
右边：\( P(A') \times P(B') = (1 - P(A)) \times (1 - P(B)) \)
因为\( P(A \cup B) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) - P(A) \times P(B) \)，故：
\( 1 - (P(A) + P(B) - P(A) \times P(B)) = 1 - P(A) - P(B) + P(A) \times P(B) = (1 - P(A)) \times (1 - P(B)) \)，相等。