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Multiplication Rule
Sections 3-4 & 3-5
M A R I O F. T R I O L A
Copyright © 1998, Triola, Elementary Statistics
Copyright © 1998, Triola, Elementary Statistics
Addison
Wesley
Longman
Addison Wesley Longman
1
Finding the Probability of
Two or More Selections
 Multiple selections
 Multiplication Rule
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Addison Wesley Longman
2
Example:
• Tom and Fred are playing a game.
One of them is asked to pick a card
from a pocket, that contains 5 cards
marked a, b, c, d, e.
Q: What is the probability of the event
A = { Tom picks the “c” card }
Copyright © 1998, Triola, Elementary Statistics
Addison Wesley Longman
3
FIGURE
Tree Diagram of Test Answers
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4
FIGURE
Tree Diagram of the events
T
F
a
b
c
d
e
a
b
c
d
e
Copyright © 1998, Triola, Elementary Statistics
Addison Wesley Longman
T&a
T&b
T&c
T&d
T&e
F&a
F&b
F&c
F&d
F&e
5
FIGURE
Tree Diagram of Test Answers
T
F
a
b
c
d
e
a
b
c
d
e
Copyright © 1998, Triola, Elementary Statistics
Addison Wesley Longman
T&a
T&b
T&c
T&d
T&e
F&a
F&b
F&c
F&d
F&e
6
FIGURE
Tree Diagram of Test Answers
T
F
P(T) =
a
b
c
d
e
a
b
c
d
e
T&a
T&b
T&c
T&d
T&e
F&a
F&b
F&c
F&d
F&e
1
2
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Addison Wesley Longman
7
FIGURE
Tree Diagram of Test Answers
T
F
P(T) =
1
2
T&a
T&b
T&c
T&d
T&e
F&a
F&b
F&c
F&d
F&e
a
b
c
d
e
a
b
c
d
e
P(c) =
1
5
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Addison Wesley Longman
8
FIGURE
Tree Diagram of Test Answers
T
F
P(T) =
1
2
T&a
T&b
T&c
T&d
T&e
F&a
F&b
F&c
F&d
F&e
a
b
c
d
e
a
b
c
d
e
P(c) =
1
5
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1
P(T and c) = 10
9
P (Tom and c card)
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10
P (Tom and c card) = P (T and c)
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P (Tom and c card) = P (T and c)
1
10
1
2
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1
5
12
P (Tom and c card) = P (T and c)
1 =
10
1
1
•
2
5
Multiplication
Rule
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13
P (Tom and c card) = P (T and c)
1 =
10
1
1
•
2
5
Multiplication
Rule
INDEPENDENT EVENTS
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Definitions
 Independent Events
Two events A and B are independent if the
occurrence of one does not affect the
probability of the occurrence of the other.
 Dependent Events
If A and B are not independent, they are
said to be dependent.
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Notation for
Multiplication Rule
P(B | A) represents the probability of B
occurring after it is assumed that event A has
already occurred (read B | A as “B given A”).
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Example: Find the probability of drawing two
cards from a shuffled deck of cards such that
the first is an Ace and the second is a King.
(The cards are drawn without replacement.)
• P(Ace on first card) =
• P(King/Ace) =
4
52
4
51
• P(drawing Ace, then a King) =
4
52
•
4
51
=
•16
.00603
•2652
DEPENDENT EVENTS
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Example: Find the probability of drawing two
cards from a shuffled deck of cards such that
the first is an Ace and the second is a King.
(The cards are drawn without replacement.)
• P(Ace on first card) =
4
52
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Addison Wesley Longman
18
Example: Find the probability of drawing two
cards from a shuffled deck of cards such that
the first is an Ace and the second is a King.
(The cards are drawn without replacement.)
• P(Ace on first card) =
• P(King
Ace) =
4
52
4
51
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Addison Wesley Longman
19
Example: Find the probability of drawing two
cards from a shuffled deck of cards such that
the first is an Ace and the second is a King.
(The cards are drawn without replacement.)
• P(Ace on first card) =
• P(King
Ace) =
4
52
4
51
• P(drawing Ace, then a King) =
Copyright © 1998, Triola, Elementary Statistics
Addison Wesley Longman
4
52
•
4
51
=
16 
0.00603
2652
20
Example: Find the probability of drawing two
cards from a shuffled deck of cards such that
the first is an Ace and the second is a King.
(The cards are drawn without replacement.)
• P(Ace on first card) =
• P(King
Ace) =
4
52
4
51
• P(drawing Ace, then a King) =
4
52
•
4
51
=
16 
0.00603
2652
DEPENDENT EVENTS
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Addison Wesley Longman
21
Formal Multiplication Rule
P(A and B) = P(A) • P(B) if A and B are independent
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Formal Multiplication Rule
P(A and B) = P(A) • P(B) if A and B are independent
P(A and B) = P(A) • P(B|A) if A and B are dependent
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Figure Applying the Multiplication Rule
P(A or B)
Multiplication Rule
Are
A and B
independent
?
Yes
P(A and B) = P(A) • P(B)
No
P(A and B) = P(A) • P(B | A)
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Intuitive Multiplication
When finding the probability that event A occurs
on one trial and B occurs on the next trial,
multiply the probability of event A by the
probability of event B, but be sure that the
probability of event B takes into account the
previous occurrence of event A.
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Small Samples
from
Large Populations
If small sample is drawn from large
population (if n  5% of N), you can treat
the events as independent.
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Conditional Probability

Definition
The conditional probability of B given A
is the probability of event B occurring,
given that A has already occurred.
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Conditional Probability
Dependent Events
P(A and B) = P(A) • P(B|A)
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Conditional Probability
Dependent Events
P(A and B) = P(A) • P(B|A)
Formal
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Conditional Probability
Dependent Events
P(A and B) = P(A) • P(B|A)
Formal
P(B|A) =
P(A and B)
P(A)
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Conditional Probability
Dependent Events
P(A and B) = P(A) • P(B|A)
Formal
P(B|A) =
Intuitive
P(A and B)
P(A)
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Conditional Probability
Dependent Events
P(A and B) = P(A) • P(B|A)
Formal
P(B|A) =
Intuitive
P(A and B)
P(A)
The conditional probability of B given A can be
found by assuming the event A has occurred and,
operating under that assumption, calculating the
probability that event B will occur.
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Probability of ‘At Least One’
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Probability of ‘At Least One’
 ‘At least one’ is equivalent to one or
more.
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Probability of ‘At Least One’
 ‘At least one’ is equivalent to one or
more.
 The complement of getting at least
one item of a particular type is that
you get no items of that type.
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Probability of ‘At Least One’
 ‘At least one’ is equivalent to one or
more.
 The complement of getting at least
one item of a particular type is that
you get no items of that type.
If P(A) = P(getting at least one), then
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Probability of ‘At Least One’
 ‘At least one’ is equivalent to one or
more.
 The complement of getting at least
one item of a particular type is that
you get no items of that type.
If P(A) = P(getting at least one), then
P(A) = 1 – P(A)
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Probability of ‘At Least One’
 ‘At least one’ is equivalent to one or
more.
 The complement of getting at least
one item of a particular type is that
you get no items of that type.
If P(A) = P(getting at least one), then
P(A) = 1 – P(A)
where P(A) is P(getting none)
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