No Title

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 E 和 E^c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(E^c) = 1 - P(E)
Proposition 2: 若 $E\subset F$ , 則 $P(E) \leq P(F)$ .
Proof:
Since $E\subset F$ , it follows that we can express F as $F=E\cup E^cF$ . Hence, as E and E^cF are mutually exclusive, we obtain from Axiom 3 that P(F)=P(E)+P(E^cF) which proves the result, since $P(E^cF)\geq 0$ $\rule[0.02em]{1.0mm}{1.5mm}$
Proposition 3: $P(E\cup F)=P(E)+P(F)-P(EF)$
Proof:
首先我們注意到 $E\cup F$ 可寫成兩個互斥事件 E 和 E^cF 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: $\displaystyle\begin{array}{rcl} P(E_1\cup E_2\cup\ldots\cup E_n)&=& \sum_{i=1}... ...E_{i2}...E_{ir})\\ \\ &+&\ldots +(-1)^{n+1}P(E_1E_2\ldots E_n) \\ \end{array}$
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

$\begin{array}{rcl} P(E\cup F\cup G)&=&P(E)+P(F)-P(EF)+P(G)-P(EG\cup FG) \\ &=&... ...G)-P(FG)+P(EGFG) \\ &=&P(E)+P(F)+P(G)-P(EF)-P(EG)-P(FG)+P(EGF) \\ \end{array}$