The Multiplication Rules

4
CHAPTER
The Multiplication
Rules
Introduction
The previous chapter showed how the addition rules could be used to solve
problems in probability. This chapter will show you how to use the
multiplication rules to solve many problems in probability. In addition, the
concept of independent and dependent events will be introduced.
56
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Independent and Dependent Events
The multiplication rules can be used to find the probability of two or more
events that occur in sequence. For example, we can find the probability of
selecting three jacks from a deck of cards on three sequential draws. Before
explaining the rules, it is necessary to differentiate between independent and
dependent events.
Two events, A and B, are said to be independent if the fact that event A
occurs does not affect the probability that event B occurs. For example, if a
coin is tossed and then a die is rolled, the outcome of the coin in no way
affects or changes the probability of the outcome of the die. Another
example would be selecting a card from a deck, replacing it, and
then selecting a second card from a deck. The outcome of the first card, as
long as it is replaced, has no effect on the probability of the outcome of the
second card.
On the other hand, when the occurrence of the first event in some way
changes the probability of the occurrence of the second event, the two events
are said to be dependent. For example, suppose a card is selected from a deck
and not replaced, and a second card is selected. In this case, the probability of
selecting any specific card on the first draw is
1

52
, but since this card is not
replaced, the probability of selecting any other specific card on the second
draw is
1
51
, since there are only 51 cards left.
Another example would be parking in a no parking zone and getting a
parking ticket. Again, if you are legally parked, the chances of getting a
parking ticket are pretty close to zero (as long as the meter does not run out).
However, if you are illegally parked, your chances of getting a parking ticket
dramatically increase.
PRACTICE
Determine whether the two events are independent or dependent.
1. Tossing a coin and selecting a card from a deck
2. Driving on ice and having an accident
3. Drawing a ball from an urn, not replacing it, and then drawing a
second ball
4. Having a high I.Q. and having a large hat size
5. Tossing one coin and then tossing a second coin
CHAPTER 4 The Multiplication Rules
57
ANSWERS
1. Independent. Tossing a coin has no effect on drawing a card.
2. Dependent. In most cases, driving on ice will increase the probability
of having an accident.
3. Dependent. Since the first ball is not replaced before the second ball is
selected, it will change the probability of a specific second ball being
selected.
4. Independent. To the best of the author’s knowledge, no studies have

been done showing any relationship between hat size and I.Q.
5. Independent. The outcome of the first coin does not influence the
outcome of the second coin.
Multiplication Rule I
Before explaining the first multiplication rule, consider the example of tossing
two coins. The sample space is HH, HT, TH, TT. From classical probability
theory, it can be determined that the probability of getting two heads is
1
4
,
since there is only one way to get two heads and there are four outcomes in
the sample space. However, there is another way to determine the probability
of getting two heads. In this case, the probability of getting a head on the first
toss is
1
2
, and the probability of getting a head on the second toss is also
1
2
.
So the probability of getting two heads can be determined by multiplying
1
2
Á
1
2
¼
1
4
This example illustrates the first multiplication rule.

Multiplication Rule I: For two independent events A and B, PðA and BÞ¼
PðAÞÁPðBÞ.
In other words, when two independent events occur in sequence, the prob-
ability that both events will occur can be found by multiplying the probabil-
ities of each individual event.
The word and is the key word and means that both events occur in
sequence and to multiply.
EXAMPLE: A coin is tossed and a die is rolled. Find the probability of
getting a tail on the coin and a 5 on the die.
SOLUTION:
Since PðtailÞ¼
1
2
and Pð5Þ¼
1
6
; Pðtail and 5Þ¼PðtailÞÁPð5Þ¼
1
2
Á
1
6
¼
1
12
.
Note that the events are independent.
CHAPTER 4 The Multiplication Rules
58
The previous example can also be solved using classical probability. Recall

that the sample space for tossing a coin and rolling a die is
H1, H2, H3, H4, H5, H6
T1, T2, T3, T4, T5, T6
Notice that there are 12 outcomes in the sample space and only one
outcome is a tail and a 5; hence, P(tail and 5) ¼
1
12
.
EXAMPLE: An urn contains 2 red balls, 3 green balls, and 5 blue balls.
A ball is selected at random and its color is noted. Then it is replaced and
another ball is selected and its color is noted. Find the probability of each of
these:
a. Selecting 2 blue balls
b. Selecting a blue ball and then a red ball
c. Selecting a green ball and then a blue ball
SOLUTION:
Since the first ball is being replaced before the second ball is selected, the
events are independent.
a. There are 5 blue balls and a total of 10 balls; therefore, the probability
of selecting two blue balls with replacement is
P(blue and blue) ¼ PðblueÞÁPðblueÞ
¼
5
10
Á
5
10
¼
25
100

¼
1
4
b. There are 5 blue balls and 2 red balls, so the probability of selecting a
blue ball and then a red ball with replacement is
Pðblue and redÞ¼PðblueÞÁPðredÞ
¼
5
10
Á
2
10
¼
10
100
¼
1
10
CHAPTER 4 The Multiplication Rules
59
c. There are 3 green balls and 5 blue balls, so the probability of selecting
a green ball and then a blue ball with replacement is
Pðgreen and blueÞ¼PðgreenÞÁPðblueÞ
¼
3
10
Á
5
10
¼

15
100
¼
3
20
The multiplication rule can be extended to 3 or more events that
occur in sequence, as shown in the next example.
EXAMPLE: A die is tossed 3 times. Find the probability of getting three 6s.
SOLUTION:
When a die is tossed, the probability of getting a six is
1
6
; hence, the probabil-
ity of getting three 6s is
Pð6 and 6 and 6Þ¼Pð6ÞÁPð6ÞÁPð6Þ
¼
1
6
Á
1
6
Á
1
6
¼
1
216
Another situation occurs in probability when subjects are selected from a
large population. Even though the subjects are not replaced, the probability
changes only slightly, so the change can be ignored. Consider the next

example.
EXAMPLE: It is known that 66% of the students at a large college favor
building a new fitness center. If two students are selected at random, find the
probability that all of them favor the building of a new fitness center.
SOLUTION:
Since the student population at the college is large, selecting a student
does not change the 66% probability that the next student selected will
favor the building of a new fitness center; hence, the probability of selecting
two students who both favor the building of a new fitness center is
(0.66)(0.66) ¼ 0.4356 or 43.56%.
CHAPTER 4 The Multiplication Rules
60
PRACTICE
1. A card is drawn from a deck, then replaced, and a second card is
drawn. Find the probability that two kings are selected.
2. If 12% of adults are left-handed, find the probability that if three
adults are selected at random, all three will be left-handed.
3. If two people are selected at random, find the probability that both
were born in August.
4. A coin is tossed 4 times. Find the probability of getting 4 heads.
5. A die is rolled and a card is selected at random from a deck of 52
cards. Find the probability of getting an odd number on the die and a
club on the card.
ANSWERS
1. The probability that 2 kings are selected is
Pðking and kingÞ¼PðkingÞÁPðkingÞ¼
4
1
52
13

Á
4
1
52
13
¼
1
169
2. The probability of selecting 3 adults who are left-handed is
(0.12)(0.12)(0.12) ¼ 0.001728.
3. Each person has approximately 1 chance in 12 of being born in
August; hence, the probability that both are born in August is
1
12
Á
1
12
¼
1
144
:
4. The probability of getting 4 heads is
1
2
Á
1
2
Á
1
2

Á
1
2
¼
1
16
.
5. The probability of getting an odd number on the die is
3
6
¼
1
2
, and the
probability of getting a club is
13
52
¼
1
4
; hence, the P(odd and
club) ¼ PðoddÞÁPðclubÞ¼
1
2
Á
1
4
¼
1
8

.
Multiplication Rule II
When two sequential events are dependent, a slight variation of the
multiplication rule is used to find the probability of both events occurring.
For example, when a card is selected from an ordinary deck of 52 cards the
CHAPTER 4 The Multiplication Rules
61
probability of getting a specific card is
1
52
, but the probability of getting a
specific card on the second draw is
1
51
since 51 cards remain.
EXAMPLE: Two cards are selected from a deck and the first card is not
replaced. Find the probability of getting two kings.
SOLUTION:
The probability of getting a king on the first draw is
4
52
and the probability
of getting a king on the second draw is
3
51
, since there are 3 kings left and
51 cards left. Hence the probability of getting 2 kings when the first card is
not replaced is
4
52

Á
3
51
¼
12
2652
¼
1
221
.
When the two events A and B are dependent, the probability that the
second event B occurs after the first event A has already occurred is written as
P(B | A). This does not mean that B is divided by A; rather, it means and is
read as ‘‘the probability that event B occurs given that event A has already
occurred.’’ P(B | A) also means the conditional probability that event B occurs
given event A has occurred. The second multiplication rule follows.
Multiplication Rule II: When two events are dependent, the probability of
both events occurring is PðA and BÞ¼PðAÞÁPðB j AÞ
EXAMPLE: A box contains 24 toasters, 3 of which are defective. If
two toasters are selected and tested, find the probability that both are
defective.
SOLUTION:
Since there are 3 defective toasters out of 24, the probability that the first
toaster is defective is
3
24
¼
1
8
. Since the second toaster is selected from the

remaining 23 and there are two defective toasters left, the probability that
it is defective is
2
23
. Hence, the probability that both toasters are defective is
PðD
1
and D
2
Þ¼PðD
1
ÞÁPðD
2
jD
1
Þ¼
3
1
24
8
4
Á
2
1
23
¼
1
92
CHAPTER 4 The Multiplication Rules
62

EXAMPLE: Two cards are drawn without replacement from a deck of
52 cards. Find the probability that both are queens.
SOLUTION:
PðQ and QÞ¼PðQÞÁPðQjQÞ
¼
4
52
Á
3
51
¼
1
221
This multiplication rule can be extended to include three or more events, as
shown in the next example.
EXAMPLE: A box contains 3 orange balls, 3 yellow balls, and 2 white balls.
Three balls are selected without replacement. Find the probability of
selecting 2 yellow balls and a white ball.
SOLUTION:
Pðyellow and yellow and whiteÞ¼
3
8
Á
2
7
Á
2
6
¼
3

1
8
4
Á
2
1
7
Á
2
1
6
1
¼
1
28
Remember that the key word for the multiplication rule is and. It means to
multiply.
When two events are dependent, the probability that the second event
occurs must be adjusted for the occurrence of the first event. For the
mathematical purist, only one multiplication rule is necessary for two events,
and that is
PðA and BÞ¼PðAÞÁPðB j AÞ:
The reason is that when the events are independent PðBjAÞ¼PðBÞ since
the occurrence of the first event A has no effect on the occurrence of the
second event B.
CHAPTER 4 The Multiplication Rules
63