Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

1.4. Conditional Probability and Independence 25

hand (A), is, sinceA∩B=B,

P(B|A)=

P(B)

P(A)

=

( 13

5

)

/

( 52

5

)

[( 13

4

)( 39

1

)

+

( 13

5

)]

/

( 52

5

)

=

( 13

5

)

( 13

4

)( 39

1

)

+

( 13

5

)=0. 0441.

Note that this is not the same as drawing for a spade to complete a flush in draw

poker; see Exercise 1.4.3.

From the definition of the conditional probability set function, we observe that

P(A∩B)=P(A)P(B|A).

This relation is frequently called themultiplication rulefor probabilities. Some-
times, after considering the nature of the random experiment, it is possible to make
reasonable assumptions so that bothP(A)andP(B|A) can be assigned. Then
P(A∩B) can be computed under these assumptions. This is illustrated in Exam-
ples 1.4.2 and 1.4.3.

Example 1.4.2. A bowl contains eight chips. Three of the chips are red and
the remaining five are blue. Two chips are to be drawn successively, at random
and without replacement. We want to compute the probability that the first draw
results in a red chip (A) and that the second draw results in a blue chip (B). It is
reasonable to assign the following probabilities:

P(A)=^38 and P(B|A)=^57.

Thus, under these assignments, we haveP(A∩B)=(^38 )(^57 )=^1556 =0.2679.

Example 1.4.3.From an ordinary deck of playing cards, cards are to be drawn

successively, at random and without replacement. The probability that the third

spade appears on the sixth draw is computed as follows. LetAbe the event of two

spades in the first five draws and letBbe the event of a spade on the sixth draw.

Thus the probability that we wish to compute isP(A∩B). It is reasonable to take

P(A)=

( 13

2

)( 39

3

)

( 52

5

) =0.2743 and P(B|A)=

11

47

=0. 2340.

The desired probabilityP(A∩B) is then the product of these two numbers, which
to four places is 0.0642.

The multiplication rule can be extended to three or more events. In the case of
three events, we have, by using the multiplication rule for two events,

P(A∩B∩C)=P[(A∩B)∩C]

= P(A∩B)P(C|A∩B).