UNIT IV: ADVANCED PROBABILITY
Conditional Probability
Discrete Math - Section 7.4
I. Warm-Up Problem
Suppose two fair dice are rolled. Find the probability of each of the following results.
(You are given a matrix of two-dice-roll outcomes for each part that you can use in figuring out your solution.)
1.
A sum of 8, given that the roll is a sum greater than 7.
11
21
31
41
51
61
2.
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
A "double," given that the sum of the dice rolled equals 9.
11
21
31
41
51
61
4.
13
23
33
43
53
63
A sum of 6, given that the roll is "doubles."
11
21
31
41
51
61
3.
12
22
32
42
52
62
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
A sum of 7, given that the roll is a sum equal to 7 or 11.
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
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II. Formalization
We return to our study of probability by considering a special class of probability called conditional probability. This kind
of problems come up when we have some given information that effectively changes the sample space.
Suppose we have two events, E and F. Suppose we are also given that event F occurs. What is the sample space for a
probability problem now?
Consider the following Venn Diagram. First shade in the sample space lightly.
E
F
U
Suppose we wanted to know the probability of E given F. What is the event? Shade the event darkly in the Venn Diagram
now.
So, using the Venn Diagram, what is the desired probability?
Notation: We use a vertical bar to mean "given." Thus, the probability of E given F is written as P(E | F).
Now that we have this notation, we can formally write a formula for conditional probability. But, not so fast… there's a
division involved, so we must not anger the math gods and divide by zero. Thus, we note the restriction that P(F) ≠ 0.
So, in summary, the formula for conditional probability is….
Theorem: Conditional Probability
Given two events E and F where P(F) ≠ 0, the probability of E given F is given by
Question: Suppose we knew P(E | F) and P(F). How could we find P(E  F) easily? Why does this make sense?
Conditional probability problems classically appear in a few different forms:
 Problems where we're given probabilities and need to apply the formula we just deduced
 Problems where we're given a table. Here, generally, columns or rows correspond to conditional sample spaces.
 Problems where we're given a possibility tree. Each branch is like a "given" here.
We shall now look at some examples.
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III. Examples
Example 1
Given P(A) = 0.65, P(B) = 0.4, and P(A  B) = 0.3, draw a Venn Diagram and use the diagram and the conditional
probability formula to find the following:
a.
P(A|B)
b.
P(B|A)
c.
P(A)

d.
P(B)

e.
P(A | B)

f.
P(B | A)

Example 2 (Freund): Research Lab
The director of a research lab has the following information: The probability that all of the equipment needed for a project
will be delivered on time is 0.75 and the probability that all equipment will be delivered on time and the project will be
completed on time is 0.6. What is the probability that the project will be completed on time, given that all of the necessary
equipment was delivered on time?
First define symbols for the events.
Now draw a Venn Diagram and apply the conditional probability formula.
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Example 3 (Freund): Elementary Student Achievement
At a certain elementary school, the probability that a student who is randomly selected will come from a two-parent home is
0.75 and the probability that he or she will come from a two-parent home and be a low achiever (gets mostly Ds and Fs) is
0.18. What is the probability that such a randomly selected student will be a low achiever given that he or she comes from a
two-parent home?
We now look at using a table to find conditional probabilities. We'll find that we don't really need a formula at all, but instead
careful reading and logic skills.
Example 4 (Freund): Car Sales Tactics
In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new car: the
aggressive approach and the passive approach. From 1160 customers the following record was kept:
Approach
Sale
No Sale
Row Total
Aggressive
270
310
580
Passive
416
164
580
Column Total
686
474
1160
Find each of the following probabilities.
a.
P ( a sale is made ) = ________
b.
P ( salesperson was aggressive ) = _______
c.
P ( a sale is made and the salesperson was aggressive ) = _________
d.
P ( a sale is made, given the fact that the salesperson was aggressive ) = ___________
Example 5: Vending Machine
A dollar-bill changer in a sort drink machine was tested with 100 $1bills. A total of 25 of the bills were counterfeit. The results of the test
are shown in the table to the right.
a.
What is the probability that a bill accepted by the changer is
legal?
b.
What is the probability that a bill is rejected given that it is legal?
c.
What is the probability that a counterfeit bill is not rejected?
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Bill
Accepted
Rejected
Legal
69
6
Counterfeit
1
24
We also consider conditional probabilities given trees.
Example 6
Find each probability by referring to the following tree diagram:
0.1
E
0.9
F
0.3
E
X
0.25
0.75
Y
0.7
F
a.
P(F)
b.
P(E)
c.
P(F|X)
d.
P(E|Y)
Homework: 7.4. #1-27 odd
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