4.1 Probability Topics1
4.1.1 Student Learning Outcomes
By the end of this chapter, the student should be able to:
• Understand and use the terminology of probability.
• Determine whether two events are mutually exclusive and whether two events are independent.
• Calculate probabilities using the Addition Rules and Multiplication Rules.
• Construct and interpret Contingency Tables.
• Construct and interpret Venn Diagrams (optional).
• Construct and interpret Tree Diagrams (optional).
4.1.2 Introduction
It is often necessary to "guess" about the outcome of an event in order to make a decision. Politicians study
polls to guess their likelihood of winning an election. Teachers choose a particular course of study based
on what they think students can comprehend. Doctors choose the treatments needed for various diseases
based on their assessment of likely results. You may have visited a casino where people play games chosen
because of the belief that the likelihood of winning is good. You may have chosen your course of study
based on the probable availability of jobs.
You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability.
Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not
to do your homework or to study for an exam, you are using probability. In this chapter, you will learn to
solve probability problems using a systematic approach.
4.1.3 Optional Collaborative Classroom Exercise
Your instructor will survey your class. Count the number of students in the class today.
• Raise your hand if you have any change in your pocket or purse. Record the number of raised hands.
• Raise your hand if you rode a bus within the past month. Record the number of raised hands.
• Raise your hand if you answered "yes" to BOTH of the first two questions. Record the number of
raised hands.
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Use the class data as estimates of the following probabilities. P(change) means the probability that a ran-
domly chosen person in your class has change in his/her pocket or purse. P(bus) means the probability that
a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.
• Find P(change).
• Find P(bus).
• Find P(change and bus) Find the probability that a randomly chosen student in your class has change
in his/her pocket or purse and rode a bus within the last month.
• Find P(change| bus) Find the probability that a randomly chosen student has change given that
he/she rode a bus within the last month. Count all the students that rode a bus. From the group
of students who rode a bus, count those who have change. The probability is equal to those who have
change and rode a bus divided by those who rode a bus.
4.2 Terminology2
Probability is a measure that is associated with how certain we are of outcomes of a particular experiment
or activity. An experiment is a planned operation carried out under controlled conditions. If the result is
not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is
an example of an experiment.
The result of an experiment is called an outcome. A sample space is a set of all possible outcomes. Three
ways to represent a sample space are to list the possible outcomes, to create a tree diagram, or to create a
Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair
coin, S = {H, T} where H = heads and T = tails are the outcomes.
An event is any combination of outcomes. Upper case letters like A and B represent events. For example,
if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an
event A is written P (A).
The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are
between 0 and 1, inclusive (includes 0 and 1 and all numbers between these values). P (A) = 0 means
the event A can never happen. P (A) = 1 means the event A always happens. P (A) = 0.5 means the
event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20
to 2,000 to 20,000 times) the relative fequency of heads approaches 0.5 (the probability of heads).
Equally likely means that each outcome of an experiment occurs with equal probability. For example, if
you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you
toss a fair coin, a Head(H) and a Tail(T) are equally likely to occur. If you randomly guess the answer to a
true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.
To calculate the probability of an event A when all outcomes in the sample space are equally likely,
count the number of outcomes for event A and divide by the total number of outcomes in the sample space.
For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T =
tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes
{HT, TH}. P (A) = 2 .
4
Suppose you roll one fair six-sided die, with the numbers {1,2,3,4,5,6} on its faces. Let event E = rolling a
number that is at least 5. There are two outcomes {5, 6}. P (E) = 2 . If you were to roll the die only a few
6
times, you would not be surprised if your observed results did not match the probability. If you were to
roll the die a very large number of times, you would expect that, overall, 2/6 of the rolls would result in an
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outcome of "at least 5". You would not expect exactly 2/6. The long-term relative frequency of obtaining
this result would approach the theoretical probability of 2/6 as the number of repetitions grows larger and
larger.
This important characteristic of probability experiments is the known as the Law of Large Numbers: as
the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment
tends to become closer and closer to the theoretical probability. Even though the outcomes don’t happen
according to any set pattern or order, overall, the long-term observed relative frequency will approach the
theoretical probability. (The word empirical is often used instead of the word observed.) The Law of Large
Numbers will be discussed again in Chapter 7.
It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may
be unfair, or biased . Two math professors in Europe had their statistics students test the Belgian 1 Euro
coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44%
of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to
draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you
have at home; the spots on each face are usually small holes carved out and then painted to make the spots
visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight
weight differences due to the different numbers of holes in the faces. Gambling casinos have a lot of money
depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice
have flat faces; the holes are completely filled with paint having the same density as the material that the
dice are made out of so that each face is equally likely to occur. Later in this chapter we will learn techniques
to use to work with probabilities for events that are not equally likely.
"OR" Event:
An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let
A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are
NOT listed twice.
"AND" Event:
An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let
A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}.
The complement of event A is denoted A’ (read "A prime"). A’ consists of all outcomes that are NOT in A.
Notice that P (A) + P (A’) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then,
A’ = {5, 6}. P (A) = 4 , P (A’) = 2 , and P (A) + P (A’) = 4 + 2 = 1
6
6
6
6
The conditional probability of A given B is written P (A|B). P (A|B) is the probability that event A will
occur given that the event B has already occurred. A conditional reduces the sample space. We calculate
the probability of A from the reduced sample space B. The formula to calculate P (A|B) is
P (A|B) = P(AANDB)
P(B)
where P (B) is greater than 0.
For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A =
face is 2 or 3 and B = face is even (2, 4, 6). To calculate P (A|B), we count the number of outcomes 2 or 3 in
the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes in B (and not S).
We get the same result by using the formula. Remember that S has 6 outcomes.
P (A|B) = P(A and B) = (the number of outcomes that are 2 or 3 and even in S) / 6 = 1/6 = 1
P(B)
(the number of outcomes that are even in S) / 6
3/6
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Understanding Terminology and Symbols
It is important to read each problem carefully to think about and understand what the events are. Under-
standing the wording is the first very important step in solving probability problems. Reread the problem
several times if necessary. Clearly identify the event of interest. Determine whether there is a condition
stated in the wording that would indicate that the probability is conditional; carefully identify the condi-
tion, if any.
Exercise 4.2.1
(Solution on p. 202.)
In a particular college class, there are male and female students. Some students have long hair and
some students have short hair. Write the symbols for the probabilities of the events for parts (a)
through (j) below. (Note that you can’t find numerical answers here. You were not given enough
information to find any probability values yet; concentrate on understanding the symbols.)
• Let F be the event that a student is female.
• Let M be the event that a student is male.
• Let S be the event that a student has short hair.
• Let L be the event that a student has long hair.
a. The probability that a student does not have long hair.
b. The probability that a student is male or has short hair.
c. The probability that a student is a female and has long hair.
d. The probability that a student is male, given that the student has long hair.
e. The probability that a student has long hair, given that the student is male.
f. Of all the female students, the probability that a student has short hair.
g. Of all students with long hair, the probability that a student is female.
h. The probability that a student is female or has long hair.
i. The probability that a randomly selected student is a male student with short hair.
j. The probability that a student is female.
**With contributions from Roberta Bloom
4.3 Independent and Mutually Exclusive Events3
Independent and mutually exclusive do not mean the same thing.
4.3.1 Independent Events
Two events are independent if the following are true:
• P (A|B) = P (A)
• P (B|A) = P (B)
• P (A AND B) = P (A) · P (B)
Two events A and B are independent if the knowledge that one occurred does not affect the chance the
other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome
of the first roll does not change the probability for the outcome of the second roll. To show two events are
independent, you must show only one of the above conditions. If two events are NOT independent, then
we say that they are dependent.
Sampling may be done with replacement or without replacement.
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• With replacement: If each member of a population is replaced after it is picked, then that member has
the possibility of being chosen more than once. When sampling is done with replacement, then events
are considered to be independent, meaning the result of the first pick will not change the probabilities
for the second pick.
• Without replacement:: When sampling is done without replacement, then each member of a popu-
lation may be chosen only once. In this case, the probabilities for the second pick are affected by the
result of the first pick. The events are considered to be dependent or not independent.
If it is not known whether A and B are independent or dependent, assume they are dependent until you
can show otherwise.
4.3.2 Mutually Exclusive Events
A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do
not share any outcomes and P(A AND B) = 0.
For
example,
suppose
the
sample
space
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Let
A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}.
A AND B
= {4, 5}.
P(A AND B) =
2 and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any
10
numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive.
If it is not known whether A and B are mutually exclusive, assume they are not until you can show other-
wise.
The following examples illustrate these definitions and terms.
Example 4.1
Flip two fair coins. (This is an experiment.)
The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The outcomes are HH,
HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin
showed heads and the second coin showed tails. The TH means that the first coin showed tails
and the second coin showed heads.
• Let A = the event of getting at most one tail. (At most one tail means 0 or 1 tail.) Then A can
be written as {HH, HT, TH}. The outcome HH shows 0 tails. HT and TH each show 1 tail.
• Let B = the event of getting all tails. B can be written as {TT}. B is the complement of A. So,
B = A’. Also, P (A) + P (B) = P (A) + P (A’) = 1.
• The probabilities for A and for B are P (A) = 3 and P (B) = 1 .
4
4
• Let C = the event of getting all heads. C = {HH}. Since B = {TT}, P (B AND C) = 0.
B and C are mutually exclusive. (B and C have no members in common because you cannot
have all tails and all heads at the same time.)
• Let D = event of getting more than one tail. D = {TT}. P (D) = 1 .
4
• Let E = event of getting a head on the first roll. (This implies you can get either a head or tail
on the second roll.) E = {HT, HH}. P (E) = 2 .
4
• Find the probability of getting at least one (1 or 2) tail in two flips. Let F = event of getting
at least one tail in two flips. F = {HT, TH, TT}. P(F) = 34
Example 4.2
Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then
A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}.
• Find the complement of A, A’. The complement of A, A’, is B because A and B together
make up the sample space. P(A) + P(B) = P(A) + P(A’) = 1. Also, P(A) = 3 and P(B) = 3
6
6
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• Let event C = odd faces larger than 2. Then C = {3, 5}. Let event D = all even faces smaller
than 5. Then D = {2, 4}. P(C and D) = 0 because you cannot have an odd and even face at
the same time. Therefore, C and D are mutually exclusive events.
• Let event E = all faces less than 5. E = {1, 2, 3, 4}.
Problem
(Solution on p. 202.)
Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?
• Find P(C|A). This is a conditional. Recall that the event C is {3, 5} and event A is {1, 3, 5}.
To find P(C|A), find the probability of C using the sample space A. You have reduced the
sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P(C|A) = 23
Example 4.3
Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking
a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are
G and H independent?
If G and H are independent, then you must show ONE of the following:
• P(G|H) = P(G)
• P(H|G) = P(H)
• P(G AND H) = P(G) · P(H)
NOTE: The choice you make depends on the information you have. You could choose any of the
methods here because you have the necessary information.
Problem 1
Show that P(G|H) = P(G).
Solution
P(G|H) = P(G AND H) = 0.3 = 0.6 = P(G)
P(H)
0.5
Problem 2
Show P(G AND H) = P(G) · P(H).
Solution
P (G) · P (H) = 0.6 · 0.5 = 0.3 = P(G AND H)
Since G and H are independent, then, knowing that a person is taking a science class does not
change the chance that he/she is taking math. If the two events had not been independent (that
is, they are dependent) then knowing that a person is taking a science class would change the
chance he/she is taking math. For practice, show that P(H|G) = P(H) to show that G and H are
independent events.
Example 4.4
In a box there are 3 red cards and 5 blue cards. The red cards are marked with the numbers 1, 2,
and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled.
You reach into the box (you cannot see into it) and draw one card.
Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.
The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has 8 outcomes.
• P(R) = 3 . P(B) = 5 . P(R AND B) = 0. (You cannot draw one card that is both red and blue.)
8
8
169
• P(E) = 3 . (There are 3 even-numbered cards, R2, B2, and B4.)
8
• P(E|B) = 2 . (There are 5 blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are
5
2 even cards: B2 and B4.)
• P(B|E) = 2 . (There are 3 even-numbered cards: R2, B2, and B4. Out of the even-numbered
3
cards, 2 are blue: B2 and B4.)
• The events R and B are mutually exclusive because P(R AND B) = 0.
• Let G = card with a number greater than 3. G = {B4, B5}. P(G) = 2 . Let H = blue card
8
numbered between 1 and 4, inclusive. H = {B1, B2, B3, B4}. P(G|H) = 1 . (The only card in
4
H that has a number greater than 3 is B4.) Since 2 = 1 , P(G) = P(G|H) which means that
8
4
G and H are independent.
Example 4.5
In a particular college class, 60% of the students are female. 50 % of all students in the class have
long hair. 45% of the students are female and have long hair. Of the female students, 75% have
long hair. Let F be the event that the student is female. Let L be the event that the student has
long hair. One student is picked randomly. Are the events of being female and having long hair
independent?
• The following probabilities are given in this example:
• P(F ) = 0.60 ; P(L ) = 0.50
• P(F AND L) = 0.45
• P(L|F) = 0.75
NOTE: The choice you make depends on the information you have. You could use the first or
last condition on the list for this example. You do not know P(F|L) yet, so you can not use the
second condition.
Solution 1
Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) =
(0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because
P(F and L) does not equal P(F)P(L).
Solution 2
check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but P(L) = 0.50; they are not
equal. The events of being female and having long hair are not independent.
Interpretation of Results
The events of being female and having long hair are not independent; knowing that a student is
female changes the probability that a student has long hair.
**Example 5 contributed by Roberta Bloom
4.4 Two Basic Rules of Probability4
4.4.1 The Multiplication Rule
If A and B are two events defined on a sample space, then: P(A AND B) = P(B) · P(A|B).
This rule may also be written as : P (A|B) = P(A AND B)
P(B)
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CHAPTER 4. PROBABILITY TOPICS
(The probability of A given B equals the probability of A and B divided by the probability of B.)
If A and B are independent, then P(A|B)
= P(A). Then P(A AND B)
= P(A|B) P(B) becomes
P(A AND B) = P(A) P(B).
4.4.2 The Addition Rule
If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) − P(A AND B).
If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) − P(A AND B)
becomes P(A OR B) = P(A) + P(B).
Example 4.6
Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B
= Alaska
• Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the
probability that he chooses B is P(B) = 0.35.
• P(A and B) = 0 because Klaus can only afford to take one vacation
• Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) =
P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go
anywhere on vacation must be 0.05.
Example 4.7
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt
two goals in a row in the next game.
A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is
successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability
that he makes the second goal GIVEN that he made the first goal is 0.90.
Problem 1
What is the probability that he makes both goals?
Solution
The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90:
P(B AND A) = P(B|A) P(A) = 0.90 ∗ 0.65 = 0.585
(4.1)
Carlos makes the