Conditional Probability
Defining Conditional Probability
The Multiplication Rule
Independence of Events
Conditional Probability
We begin with a simple example:
 Toss a balanced die once and record the number
on the top face.
 Let E be the event that a 1 shows on the top
face.
 Let F be the event that the number on the top
face is odd.
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
What is P(E)?
What is P(E) if we are told that the number on the top
face is odd, that is, we know that the event F has
occurred?
Conditional Probability
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Key idea: The original sample space no
longer applies.
The new or reduced sample space is
S={1, 3, 5}
Notice that the new sample space
consists only of the outcomes in F.
P(E occurs given that F occurs) = 1/3
Notation: P(E|F) = 1/3
Conditional Probability
Definition. If A and B are events and
P(B)≠0, then we define the conditional
probability of A given B by
P( A  B)
P( A B) 
P( B)
Conditional Probability
P( A  B)
P( A B) 
P( B)
A
S
B
Example One
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Suppose that a balanced die is tossed once.
The relevant sample space is
S={1, 2, 3, 4, 5, 6}
Let E be the event that we observe a 1. Since
each element in S has an equal probability of
occurring we have the unconditional probability
n( E ) 1
P( E ) 

n( S ) 6
Example One

Define O to be the event that we observe an
odd number. Then the unconditional probability
is given by
P(O) 

n(O) 3 1
 
n( S ) 6 2
Now, what is the probability of E given that O
has occurred? Or, we might ask, what is the
probability of O given that E has occurred?
Example One
Symbolically, we write
P ( E | O ) or P (O | E )
to represent the two previous questions.

In both cases, our original sample space
S is not longer relevant. Let’s examine
each case separately.
Example One
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Consider P ( E | O )
The new sample space is S ’={1,3,5}
So,
n( E ) 1
P( E | O) 

n(O) 3
Example One

Using the definition we find
P( E  O) 1 / 6 1
P( E | O) 


P(O)
3/ 6 3

Similarly, we have
P(O  E ) 1 / 6
P(O | E ) 

1
P( E )
1/ 6
Conditional Probability
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Conditional probability problems lend
themselves to several methods of
solution. We can
Apply the definition
Construct Venn Diagrams
Construct Frequency (Contingency)
Tables
Construct Tree Diagrams
Example Two
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The probability that event A occurs is .63.
The probability that event B occurs is .45.
The probability that both events occur is
.10.
Find the following probabilities by using the
definition and Venn Diagrams:
a. P(A|B); b. P(B|A); c. P(AC|B)
d. P(BC|A); e. P(AC|BC)
Example Three – Using Tables
Consider the experiment of rolling a fair
die twice.
All outcomes in S are equally likely.
 1,1

1, 2 
1,3
S 
1, 4 
1,5 

1, 6 
 2,1
 2, 2 
 2,3
 2, 4 
 2,5
 2, 6 
 3,1
 3, 2 
 3,3
 3, 4 
 3,5
 3, 6 
 4,1
 4, 2 
 4,3
 4, 4 
 4,5
 4, 6 
 5,1
 5, 2 
 5,3
 5, 4 
 5,5
 5, 6 
 6,1 
 6, 2 
 6,3 

6,
4
 
 6,5 

 6, 6  
Example Three
 1,1

1, 2 
1,3
S 
1, 4 
1,5 

1,6 
 2,1
 2, 2 
 2,3
 2, 4 
 2,5
 2,6 
 3,1
 3, 2 
 3,3
 3, 4 
 3,5
3,6 
 4,1
 4, 2 
 4,3
 4, 4 
 4,5
 4,6 
 5,1
5, 2 
 5,3
5, 4 
 5,5
5,6 
 6,1 
 6, 2 
 6,3 
 6, 4 
 6,5 

 6,6 
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Let E be the event that
the sum of the faces is
even
Let T be the event that
the second die is a 2
Find
1. P(T | E)
2. P(E |T)
Example Three
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One way of doing this is to construct a table
of frequencies:
Event Ac
Event A
A B
Event B
Event B
c
A B
Total A
C
TOTALS
A  B Total B
C
AC  BC
Total Ac
Total Bc
Grand Total
Example Three
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Let’s try it to find P(T | E) and P(E | T)
Event E
Event Ec
Event T
Event T c
Remember: P( E ) 
# of successes
Total # of possible outcomes
TOTALS
Multiplication Rule
Using our definition of conditional probability
P( A  B)
P( A | B) 
P( B)
we can perform the simple algebra of
multiplying both sides of the equation by P(B)
to get
P( A  B)  P( A | B) P( B)
This is known as the multiplication rule of
probabilities and provides another way at
getting to the joint probability of A and B.
Example Four – Using Trees
If a fair coin is flipped three times, what
is the probability that it comes up tails at
least once given:
1. No information at all.
2. All three coins produce the same side.
3. It comes up tails at most once.
4. The third flip is heads.
Example Five – Trees
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Three manufacturing plants A, B, and C supply 20, 30
and 50%, respectively of all shock absorbers used by
a certain automobile manufacturer. Records show
that the percentage of defective items produced by
A, B and C is 3, 2 and 1%, respectively. What is the
probability that a randomly chosen shock absorber
installed by the manufacturer will be defective?
What is the probability that the part came from
manufacturer A, given that the part was defective?
What is the probability that the part came from B,
given that the part was not defective?
Independent Events
If the probability of the occurrence of event A
is the same regardless of whether or not an
outcome B occurs, then the outcomes A and
B are said to be independent of one another.
Symbolically, if
P( A | B)  P( A)
then A and B are independent events.
Independent Events
Recalling that P( A  B)  P( A | B) P( B)
then we can also state the following
relationship for independent events:
P( A  B)  P( A) P( B)
if and only if A and B are independent
events.
Example Six

A fair coin is tossed twice. Find the
probability that
a. the second toss is a head;
b. the second toss is a head given that
the first toss is a head.
Example Seven
A card is drawn from a standard deck of 52
cards and then a second card is drawn
without replacing the first card. Let A be
the event that a second card is a spade
and let B be the event that the first card is
a spade. Determine whether A and B are
independent events.
Example Eight
You throw two fair die, one green and one
red and observe the numbers. Decide which
pairs of events, A and B, are independent:
1. A : the sum is 5
B : the red die shows a 2
2. A : the sum is 5
B : the sum is less than 4
3. A : the sum is even
B : the red die is even
Independence
The notion of independent events can
be extended to conditional probabilities.
If E, F, and G are three events, then E
and F are independent, given that G
has happened, if
P( E  F | G )  P( E | G )  P( F | G )
Independence
Likewise, events E, F, G are independent,
given that an event H has happened, if
P( E  F  G | H )  P( E | H )  P( F | H )  P(G | H )
P( E  F | H )  P( E | H )  P( F | H )
P( F  G | H )  P( F | H )  P(G | H )
P( E  G | H )  P( E | H )  P(G | H )
Example Nine
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In the manufacture of light bulbs, filaments,
glass casings and bases are manufactured
separately and then assembled into the final
product. Past records show: 2% of filaments
are defective, 3% of casings are defective
and 1% of bases are defective.
What is the probability that one bulb chosen
at random will have no defects?