CHAP. 7] PROBABILITY 129
Multiplying both sides byP (A)gives us the following useful result:
Theorem 7.6 (Multiplication Theorem for Conditional Probability):
P(A∩B)=P (A)P (B|A)
The multiplication theorem gives us a formula for the probability that eventsAandBboth occur. It can
easily be extended to three or more eventsA 1 ,A 2 ,...Am; that is,
P(A 1 ∩A 2 ∩···Am)=P(A 1 )·P(A 2 |A 1 )···P(Am|A 1 ∩A 2 ∩···∩Am− 1 )
EXAMPLE 7.8A lot contains 12 items of which 4 are defective. Three items are drawn at random from the lot
one after the other. Find the probabilitypthat all three are nondefective.
The probability that the first item is nondefective is 128 since 8 of 12 items are nondefective. If the first item is
nondefective, then the probability that the next item is nondefective is 117 since only 7 of the remaining 11 items
are nondefective. If the first 2 items are nondefective, then the probability that the last item is nondefective is 106
since only 6 of the remaining 10 items are now nondefective. Thus by the multiplication theorem,
p=
8
12
·
7
11
·
6
10
=
14
55
≈ 0. 25
7.5Independent Events
EventsAandBin a probability spaceSare said to beindependentif the occurrence of one of them does not
influence the occurrence of the other. More specifically,Bis independent ofAifP(B)is the same asP(B|A).
Now substitutingP(B)forP(B|A)in the Multiplication TheoremP(A∩B)=P (A)P (B|A)yields
P(A∩B)=P (A)P (B).
We formally use the above equation as our definition of independence.
Definition 7.2: EventsAandBareindependentifP(A∩B)=P(A)P(B); otherwise they aredependent.
We emphasize that independence is a symmetric relation. In particular, the equation
P(A∩B)=P (A)P (B) implies both P(B|A)=P(B) and P(A|B)=P (A)
EXAMPLE 7.9 A fair coin is tossed three times yielding the equiprobable space
S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
Consider the events:
A={first toss is heads}={HHH,HHT,HTH,HTT}
B={secondtossis heads)={HHH,HHT,THH,THT}
C={exactly two heads in a row}={HHT,THH}
ClearlyAandBare independent events; this fact is verified below. On the other hand, the relationship between
AandCand betweenBandCis not obvious. We claim thatAandCare independent, but thatBandCare
dependent. We have:
P (A)=^48 =^12 , P (B)=^48 =^12 , P (C)=^28 =^14