Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

30 Probability and Distributions

In particular, ifA 1 ,A 2 ,...,Anare mutually independent, then

P(A 1 ∩A 2 ∩···∩An)=P(A 1 )P(A 2 )···P(An).

Also, as with two sets, many combinations of these events and their complements
are independent, such as

1. The eventsAc 1 andA 2 ∪Ac 3 ∪A 4 are independent,


2. The eventsA 1 ∪Ac 2 ,Ac 3 andA 4 ∩Ac 5 are mutually independent.

If there is no possibility of misunderstanding,independentis often used without the

modifiermutuallywhen considering more than two events.

Example 1.4.9.Pairwise independence does not imply mutual independence. As
an example, suppose we twice spin a fair spinner with the numbers 1, 2, 3, and 4.
LetA 1 be the event that the sum of the numbers spun is 5, letA 2 be the event that
the first number spun is a 1, and letA 3 be the event that the second number spun
is a 4. ThenP(Ai)=1/4,i=1, 2 ,3, and fori =j,P(Ai∩Aj)=1/16. So the
three events are pairwise independent. ButA 1 ∩A 2 ∩A 3 is the event that (1,4) is
spun, which has probability 1/ 16
=1/64 =P(A 1 )P(A 2 )P(A 3 ). Hence the events
A 1 ,A 2 ,andA 3 are not mutually independent.

We often perform a sequence of random experiments in such a way that the
events associated with one of them are independent of the events associated with
the others. For convenience, we refer to these events as as outcomes ofindependent
experiments, meaning that the respective events are independent. Thus we often
refer to independent flips of a coin or independent casts of a die or, more generally,
independent trials of some given random experiment.

Example 1.4.10.A coin is flipped independently several times. Let the eventAi
represent a head (H) on theith toss; thusAcirepresents a tail (T). Assume thatAi
andAciare equally likely; that is,P(Ai)=P(Aci)=^12. Thus the probability of an
ordered sequence like HHTH is, from independence,

P(A 1 ∩A 2 ∩Ac 3 ∩A 4 )=P(A 1 )P(A 2 )P(Ac 3 )P(A 4 )=(^12 )^4 = 161.

Similarly, the probability of observing the first head on the third flip is

P(Ac 1 ∩Ac 2 ∩A 3 )=P(Ac 1 )P(Ac 2 )P(A 3 )=(^12 )^3 =^18.

Also, the probability of getting at least one head on four flips is

P(A 1 ∪A 2 ∪A 3 ∪A 4 )=1−P[(A 1 ∪A 2 ∪A 3 ∪A 4 )c]

=1−P(Ac 1 ∩Ac 2 ∩Ac 3 ∩Ac 4 )

=1−

( 1

2

) 4

=^1516.

See Exercise 1.4.13 to justify this last probability.