Some Simple Propositions

在本節中,我們將提到有關機率的一些簡單命題. 首先, 我們注意到 EEc永遠是 互斥的, 而且因 $E\cup E^c=S $, 所以由 Axiom 2 和 3 得

$1=P(S)=P(E\cup E^c)=P(E)+P(E^c)$

Proposition 1
P(Ec) = 1 - P(E)
Proposition 2
$E\subset F $, 則 $P(E) \leq P(F)$.
Since $E\subset F $, it follows that we can express F as $F=E\cup E^cF$. Hence, as E and EcF are mutually exclusive, we obtain from Axiom 3 that P(F)=P(E)+P(EcF) which proves the result, since $P(E^cF)\geq 0$          $\rule[0.02em]{1.0mm}{1.5mm}$
Proposition 3
$P(E\cup F)=P(E)+P(F)-P(EF)$
首先我們注意到 $E\cup F$ 可寫成兩個互斥事件 EEcF 的聯集. 因此, 由 Axiom 3 我們得到 $P(E \cup F) = P(E \cup E^cF) = P(E) + P(E^cF)$又因 $P(E^cF)\geq 0$, 故命題得證.          $\rule[0.02em]{1.0mm}{1.5mm}$
Proposition 4
P(E_1\cup E_2\cup\ldots\cup E_n)&=&
...E_{i2}...E_{ir})\\ \\
&+&\ldots +(-1)^{n+1}P(E_1E_2\ldots E_n) \\
其中 $\displaystyle\sum_{i1<i2<\ldots<ir} P(E_{i1}E_{i2}\ldotsE_{ir})$ 是對集合 ${1,2,\ldots,n}$ 中元素個數為 r 的所有 ${n\choose r}$ 個可能子集合求其總和的.

We may also calculate the probability that any one of the three events E or F or G occurs: $P(E\cup F\cup G)=P[(E\cup F)\cup G]$ which by Proposition 3 equals $P(E\cup F)+P(G)-P[(E\cup F)G]$. Now it follows from the distributive law that the events $(E\cup F)G$ and $EG\cup FG$ are equivalent, and hence we obtain from the preceding equations that

P(E\cup F\cup G)&=&P(E)+P(F)-P(EF)+P(G)-P(EG\cup FG) \\
...G)-P(FG)+P(EGFG) \\
&=&P(E)+P(F)+P(G)-P(EF)-P(EG)-P(FG)+P(EGF) \\