抛两枚硬币的样本空间是 S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}S={HH,HT,TH,TT},定义事件:
用德摩根定律简化:
已知某股票市场日行情概率:
在样本空间 S={a,b,c}S = \{a,b,c\}S={a,b,c} 中,以下概率分配是否满足柯尔莫哥洛夫公理?
设样本空间 S={1,2,3,4,5}S = \{1,2,3,4,5\}S={1,2,3,4,5},已知:
(a) E∪F={HH,HT,TH,TT}=SE \cup F = \{HH, HT, TH, TT\} = SE∪F={HH,HT,TH,TT}=S
(b) E∩F={HT,TT}E \cap F = \{HT, TT\}E∩F={HT,TT}
(c) Fc={HH,HT}F^c = \{HH, HT\}Fc={HH,HT}
解析:F是第二次为反面的结果{HT,TT}\{HT, TT\}{HT,TT},其补集为第二次不是反面的结果
(A∪B∪C)c=Ac∩Bc∩Cc \left( A \cup B \cup C \right)^c = A^c \cap B^c \cap C^c (A∪B∪C)c=Ac∩Bc∩Cc
解析:根据德摩根第一定律的扩展形式
P(不上涨∪成交量≤1亿)=1−P(上涨∩成交量>1亿)=1−0.3=0.7P(\text{不上涨} \cup \text{成交量≤1亿}) = 1 - P(\text{上涨} \cap \text{成交量>1亿}) = 1 - 0.3 = 0.7P(不上涨∪成交量≤1亿)=1−P(上涨∩成交量>1亿)=1−0.3=0.7
技巧:利用 P(A∪B)=1−P(Ac∩Bc)P(A \cup B) = 1 - P(A^c \cap B^c)P(A∪B)=1−P(Ac∩Bc)
不满足公理:
(a) 由规范性公理:
0.2+0.2+0.1+0.1+P({5})=1⇒P({5})=0.40.2 + 0.2 + 0.1 + 0.1 + P(\{5\}) = 1 ⇒ P(\{5\}) = 0.40.2+0.2+0.1+0.1+P({5})=1⇒P({5})=0.4
(b) P(奇数)=P({1})+P({3})+P({5})=0.2+0.1+0.4=0.7P(\text{奇数}) = P(\{1\}) + P(\{3\}) + P(\{5\}) = 0.2 + 0.1 + 0.4 = 0.7P(奇数)=P({1})+P({3})+P({5})=0.2+0.1+0.4=0.7
注意:奇数集合是{1,3,5}\{1,3,5\}{1,3,5}