26 Probability and DistributionsButP(A∩B)=P(A)P(B|A). Hence, providedP(A∩B)>0,P(A∩B∩C)=P(A)P(B|A)P(C|A∩B).This procedure can be used to extend the multiplication rule to four or more
events. The general formula forkevents can be proved by mathematical induction.
Example 1.4.4.Four cards are to be dealt successively, at random and without
replacement, from an ordinary deck of playing cards. The probability of receiving a
spade, a heart, a diamond, and a club, in that order, is (^1352 )(^1351 )(^1350 )(^1349 )=0.0044.
This follows from the extension of the multiplication rule.
Considerkmutually exclusive and exhaustive eventsA 1 ,A 2 ,...,Aksuch that
P(Ai)>0,i=1, 2 ,...,k; i.e.,A 1 ,A 2 ,...,Akform a partition ofC. Here the events
A 1 ,A 2 ,...,Akdonotneed to be equally likely. LetBbe another event such that
P(B)>0. ThusBoccurs with one and only one of the eventsA 1 ,A 2 ,...,Ak;that
is,
B = B∩(A 1 ∪A 2 ∪···Ak)
=(B∩A 1 )∪(B∩A 2 )∪···∪(B∩Ak).SinceB∩Ai,i=1, 2 ,...,k, are mutually exclusive, we have
P(B)=P(B∩A 1 )+P(B∩A 2 )+···+P(B∩Ak).However,P(B∩Ai)=P(Ai)P(B|Ai),i=1, 2 ,...,k;so
P(B)=P(A 1 )P(B|A 1 )+P(A 2 )P(B|A 2 )+···+P(Ak)P(B|Ak)=∑ki=1P(Ai)P(B|Ai). (1.4.2)This result is sometimes called thelaw of total probabilityand it leads to the
following important theorem.
Theorem 1.4.1(Bayes).LetA 1 ,A 2 ,...,Akbe events such thatP(Ai)> 0 ,i=
1 , 2 ,...,k. Assume further thatA 1 ,A 2 ,...,Akform a partition of the sample space
C.LetBbe any event. Then
P(Aj|B)=P(Aj)P(B|Aj)
∑k
i=1P(Ai)P(B|Ai), (1.4.3)Proof:Based on the definition of conditional probability, we haveP(Aj|B)=P(B∩Aj)
P(B)=P(Aj)P(B|Aj)
P(B).The result then follows by the law of total probability, (1.4.2).
This theorem is the well-knownBayes’ Theorem. This permits us to calculate
the conditional probability ofAj,givenB, from the probabilities ofA 1 ,A 2 ,...,Ak
and the conditional probabilities ofB,givenAi,i=1, 2 ,...,k. The next three
examples illustrate the usefulness of Bayes Theorem to determine probabilities.