14 Probability and DistributionsThat these identities hold for all setsAandBfollows from set theory. Also, the
Venn diagrams of Figure 1.3.1 offer a verification of them.
Thus, from (3) of Definition 1.3.1,P(A∪B)=P(A)+P(Ac∩B)and
P(B)=P(A∩B)+P(Ac∩B).
If the second of these equations is solved forP(Ac∩B) and this result is substituted
in the first equation, we obtain
P(A∪B)=P(A)+P(B)−P(A∩B).This completes the proof.
Panel (a)
A∪B=A∪(Ac∩B)
AB
A=(A∩Bc)∪(A∩B)ABPanel (b)
Figure 1.3.1:Venn diagrams depicting the two disjoint unions given in expression
(1.3.1). Panel (a) depicts the first disjoint union while Panel (b) shows the second
disjoint union.Example 1.3.1.LetCdenote the sample space of Example 1.1.2. Let the proba-
bility set function assign a probability of 361 to each of the 36 points inC;thatis,the
dice are fair. IfC 1 ={(1,1),(2,1),(3,1),(4,1),(5,1)}andC 2 ={(1,2),(2,2),(3,2)},
thenP(C 1 )= 365 ,P(C 2 )= 363 ,P(C 1 ∪C 2 )= 368 ,andP(C 1 ∩C 2 )=0.
Example 1.3.2.Two coins are to be tossed and the outcome is the ordered pair
(face on the first coin, face on the second coin). Thus the sample space may be
represented asC={(H, H),(H, T),(T,H),(T,T)}. Let the probability set function
assign a probability of^14 to each element ofC.LetC 1 ={(H, H),(H, T)}and
C 2 ={(H, H),(T,H)}.ThenP(C 1 )=P(C 2 )=^12 ,P(C 1 ∩C 2 )=^14 , and, in
accordance with Theorem 1.3.5,P(C 1 ∪C 2 )=^12 +^12 −^14 =^34.