Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

1.4. Conditional Probability and Independence 29

Definition 1.4.2.LetAandBbe two events. We say thatAandBareinde-

pendentifP(A∩B)=P(A)P(B).

SupposeAandBare independent events. Then the following three pairs of
events are independent:AcandB,AandBc,andAcandBc. We show the first
and leave the other two to the exercises; see Exercise 1.4.11. Using the disjoint
union,B=(Ac∩B)∪(A∩B), we have

P(Ac∩B)=P(B)−P(A∩B)=P(B)−P(A)P(B)=[1−P(A)]P(B)=P(Ac)P(B).
(1.4.6)
Hence,AcandBare also independent.

Remark 1.4.1.Events that areindependentare sometimes calledstatistically in-
dependent,stochastically independent,orindependent in a probability sense.In
most instances, we useindependentwithout a modifier if there is no possibility of
misunderstanding.

Example 1.4.8.A red die and a white die are cast in such a way that the numbers

of spots on the two sides that are up are independent events. IfArepresents a

four on the red die andBrepresents a three on the white die, with an equally

likely assumption for each side, we assignP(A)=^16 andP(B)=^16 .Thus,from

independence, the probability of the ordered pair (red = 4, white = 3) is

P[(4,3)] = (^16 )(^16 )= 361.

The probability that the sum of the up spots of the two dice equals seven is

P[(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)]

=

( 1

6

)( 1

6

)

+

( 1

6

)( 1

6

)

+

( 1

6

)( 1

6

)

+

( 1

6

)( 1

6

)

+

( 1

6

)( 1

6

)

+

( 1

6

)( 1

6

)

= 366.

In a similar manner, it is easy to show that the probabilities of the sums of the

upfaces 2, 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 ,12 are, respectively,

1

36 ,

2

36 ,

3

36 ,

4

36 ,

5

36 ,

6

36 ,

5

36 ,

4

36 ,

3

36 ,

2

36 ,

1
36.
Suppose now that we have three events,A 1 ,A 2 ,andA 3. We say that they are
mutually independentif and only if they arepairwise independent:

P(A 1 ∩A 3 )=P(A 1 )P(A 3 ),P(A 1 ∩A 2 )=P(A 1 )P(A 2 ),

P(A 2 ∩A 3 )=P(A 2 )P(A 3 ),

and
P(A 1 ∩A 2 ∩A 3 )=P(A 1 )P(A 2 )P(A 3 ).

More generally, theneventsA 1 ,A 2 ,...,Anaremutually independentif and only
if for every collection ofkof these events, 2≤k≤n, and for every permutation
d 1 ,d 2 ,...,dkof 1, 2 ,...,k,

P(Ad 1 ∩Ad 2 ∩···∩Adk)=P(Ad 1 )P(Ad 2 )···P(Adk).