Introduction to
Probability
Introduction to Probability
• A probability experiment (or trial) is
any process with a result determined
by chance.
• Each individual result that is possible
for a probability experiment is an
outcome.
• The sample space is the set of all
possible outcomes for a given
probability experiment.
• An event is a subset of outcomes from
the sample space.
Identifying Outcomes in a Sample
Space or Event
• Consider an experiment in which a coin is
tossed and then a six-sided die is rolled.
a. List the outcomes in the sample space
for the experiment.
Identifying Outcomes in a Sample Space or Event
(cont.)
Solution
a. Each outcome consists of a coin toss and a die
roll. For example, heads and a 3 could be
denoted as H3. Using this notation, the sample
space can be written as follows. H1 T1
H2

H3
Sample space  
H4

H5
H6
T2

T3

T4

T5
T6
Identifying Outcomes in a Sample Space or Event
(cont.)
Consider an experiment in which a coin is
tossed and then a six-sided die is rolled.
b. List the outcomes in the event “tossing a
tail then rolling an odd number.”
Identifying Outcomes in a Sample Space or
Event (cont.)
b. Choosing the members of the sample
space that fit the event “tossing a tail then
rolling an odd number” gives the following:
 T1, T3, T5
Identifying Outcomes in a Sample
Space
• Consider the experiment in which two die are
rolled together. List the outcomes in the
sample space.
.
Using a Pattern to List All Outcomes in a
Sample Space
The previous question has the same
answer as the following:
Consider the experiment in which a red sixsided die and a blue six-sided die are rolled
together.
a. Use a pattern to help list the outcomes
in the sample space.
Using a Pattern to List All Outcomes in
a Sample Space (cont.)
Solution
Although there are many patterns
that could help us list all outcomes in
the sample space, one pattern will be
to start with keeping the red die at 1,
and allowing the blue die to vary from
1 to 6.
a.
Using a Pattern to List All Outcomes in
a Sample Space (cont.)
• There are six of these outcomes:
•
and
.
• Next we list all outcomes in which a 2 is rolled on the
red die. Again, there are six outcomes of this form.
The pattern continues until all 36 outcomes are
listed.
Using a Pattern to List All Outcomes in
a Sample Space (cont.)
• Sample space =
Using a Pattern to List All Outcomes in
a Sample Space (cont.)
• Note that rolling a 1 on the red die and a 2 on
the blue die, denoted
, is a different
outcome than rolling a 2 on the red die and a
1 on the blue die, denoted
.
Using a Pattern to List All Outcomes in a Sample
Space (cont.)
Consider the experiment in which a
red six-sided die and a blue six-sided
die are rolled together.
b. List the outcomes in the event “the
sum of the numbers rolled on the two
dice equals 6.”
Using a Pattern to List All Outcomes in
a Sample Space (cont.)
b. Note that 1 + 5 = 6, so the outcome of rolling a 1 on
the red die and a 5 on the blue die is in the event. The
reverse would be rolling a 5 on the red die and a 1 on the
blue die. These are two separate possible outcomes that
are both in the event. The event consists of the following
outcomes.
Using a Tree Diagram to List All Outcomes in
a Sample Space
Consider a family with three children. Use a tree
diagram to find the sample space for the gender
of each child in regard to birth order.
Using a Tree Diagram to List All Outcomes in
a Sample Space
Consider a family with three children. Use a tree
diagram to find the sample space for the gender
of each child in regard to birth order.
Solution
The tree begins with the two possibilities for the
first child—girl or boy. It then branches for each
of the other two births in the family as shown.
Using a Tree Diagram to List All Outcomes in a Sample
Space (cont.)
Key: The
notation GBG
indicates the
family shown in
red with a girl,
then a boy, and
finally a girl.
Using a Tree Diagram to List All Outcomes in a Sample
Space (cont.)
• Using the tree diagram as a guide, the sample
space can be written as follows.
GGG GGB GBG GBB
Sample space  

BGG BGB BBG BBB 
Types of Probability
Experimental probability (or empirical probability)
uses the outcomes obtained by repeatedly
performing an experiment to calculate the
probability.
Classical probability (or theoretical probability) is
the most precise type of probability and can only
be calculated when all possible outcomes in the
sample space are known and equally likely to occur.
Introduction to Probability
Experimental Probability
• In experimental probability, if E is an
event, then P(E), read “the probability
that E occurs,” is given by
f
P E  
n
• where f is the frequency of event E
and n is the total number of times the
experiment is performed.
Introduction to Probability
• Classical Probability
• In classical probability, if all outcomes are
equally likely to occur, then P(E), read “the
probability that E occurs,” is given by
nE 
P E  
n S 
• where n(E) is the number of outcomes in
the event and
• n(S) is the number of outcomes in the
sample space.
Calculating Classical Probability
Beck is allergic to peanuts. At a large dinner party one
evening, he notices that the cheesecake options on the
dessert table contain the following flavors: 10 slices of
chocolate, 12 slices of caramel, 12 slices of peanut
butter chocolate, and 8 slices of strawberry. Assume
that the desserts are served to guests at random.
a.
What is the probability that Beck’s cheesecake
contains peanuts?
Calculating Classical Probability (cont.)
• Solution
a. There are 12 slices of chocolate peanut butter
cheesecake, and 10 + 12 + 12 + 8 = 42 pieces of
cheesecake total. The probability is then
calculated as follows.
nE 
P peanut butter chocolate 
n S 
12 2
   0.2857
42 7
Calculating Classical Probability (cont.)
Beck is allergic to peanuts. At a large dinner party one
evening, he notices that the cheesecake options on the
dessert table contain the following flavors: 10 slices of
chocolate, 12 slices of caramel, 12 slices of peanut
butter chocolate, and 8 slices of strawberry. Assume
that the desserts are served to guests at random.
b. What is the probability that Beck’s dessert does not
contain chocolate?
Calculating Classical Probability (cont.)
b. There are 12 + 8 = 20 slices of cheesecake that
do not contain chocolate. The probability of
being served one of these desserts is calculated
as follows.
nE 
P not chocolate 
n S
20 10
   0.4762
42 21
Calculating Classical Probability
Consider a beginning archer who only manages
to hit the target 50% of the time. What is the
probability that in three shots, the archer will hit
the target all three times?
Calculating Classical Probability
(cont.)
• Let’s use a tree diagram.
Calculating Classical Probability (cont.)
• Notice that this gives 8 possible outcomes for
the three shots, only 1 of which consists of
hitting the target all three times. Thus, the
probability of the novice archer hitting the
target three times in a row is calculated as
nE 
follows.
P  all three hits  
n S
1

8
 0.125
Calculating Classical Probability
Consider a family with six boys. What
is the probability that the seventh
child will also be a boy?
Example 4.7: Calculating Classical
Probability (cont.)
• The only two outcomes left in the sample space
would be BBBBBBG and BBBBBBB. The
probability of baby number seven being a boy is
then
nE 
P  all seven boys  
n S 
1

2
 0.5,
• just as with any other pregnancy!
Calculating Classical Probability
In biology, we learn that many diseases are genetic. One
example of such a disease is Huntington’s disease, which
causes neurological disorders as a person ages. Each person
has two huntington genes—one inherited from each
parent. If an individual inherits a mutated huntington gene
from either of his or her parents, he or she will develop the
disease. On the TV show House, the character who Dr.
House calls “13” inherited this disease from her mother.
Assume for a moment that “13” has a child with a person
who has two healthy huntington genes. What is the
probability that her child will develop Huntington’s disease?
Calculating Classical Probability (cont.)
So the four possibilities for the child’s genes are
{Hh, Hh, hh, hh}. Someone only has to
have one mutated gene to develop the
disease, so that means two of the four
combinations would result in a child
having the disease, {Hh, Hh}. Thus, if
E = the event of the child inheriting
a mutated huntington gene, then the probability that E
occurs is calculated as follows.
Calculating Classical Probability (cont.)
nE 
P E  
n S
2 1
   0.5
4 2
This means that “13” has a 50% chance of
passing on the disease to a child. This is true
for anyone with Huntington’s disease. The
probability goes up if the other parent also
has the disease.
More Probability
Properties of Probability
• Properties of Probability
• 1. For any event, E, 0  P  E   1.
• 2. For any sample space S, P(S) = 1.
• 3. For the empty event ∅, P(∅) = 0.
The Complement
Complement Rule for Probability
The sum of the probabilities of an event, E, and
its complement, Ec , is equal to one.
 
P E   P E c  1
Using the Complement Rule for
Probability (cont.)
b. If there is a 5% chance that none of the items
on a scratch-off lottery ticket will be a winner,
what is the probability that at least one of the
scratch-off items will win?
Using the Complement Rule for
Probability (cont.)
b. The complement to having none of something is
having at least one of that thing. Thus, the
probability is calculated as follows.
P  at least one winner   1  P no winners 
 1  0.05
 0.95
 95%
• Thus, there is a 95% chance of having at least
one winning scratch-off item.
Using the Complement Rule for
Probability
• Roll a pair of standard six-sided dice. What is
the probability that neither die is a three?
Using the Complement Rule for
Probability (continued)
• Solution
We could list the outcomes in E, every combination
of the dice that does not have a three. It is much
easier to count the outcomes in the complement,
Ec. The complement of this event contains the
outcomes in which either die is a three. (Check for
yourself by making sure that adding the event and
its complement covers the entire sample space.)
Let’s list these outcomes.
Using the Complement Rule for
Probability (cont.)
• Let’s list these outcomes.
Ec =
Using the Complement Rule for
Probability (cont.)
• There are 11 outcomes where at least one of
the dice is a three. Since we have already seen
that there are 36 possible ways to roll a pair of
dice we have that


11
P E  
 .
n  S  36
c
nE
c
Using the Complement Rule for
Probability (cont.)
• Subtracting this probability from 1 gives us the
following.
P E   1  P E c 
11
1
36
25

36
 0.6944
• Therefore, the probability that neither die is a
three is approximately 0.6944.
Using the Addition Rule for Probability
Teresa is looking for a new condo to rent. Teresa’s
realtor has provided her with the following list of
amenities for 17 available properties. The list
contains the following.
• Close to the subway: 6 properties
• Low maintenance fee: 7 properties
• Green space: 5 properties
• Newly renovated: 2 properties
• Close to the subway and low maintenance fee:
2 properties
• Green space and newly renovated: 1 property
Using the Addition Rule for Probability
(cont.)
• If Teresa’s realtor selects the first condo they
visit at random, what is the probability that
the property is either close to the subway or
has a low maintenance fee?
Using the Addition Rule for Probability
(cont.)
• Solution
Before we calculate the probability, let’s verify that
Teresa’s realtor has accurately counted the total
number of properties. At first glance, it might seem
that there are more than 17 properties if you simply
add all the numbers in the list the realtor gave.
However, there are 6 + 7 + 5 + 2 = 20 properties
that have single characteristics, and 2 + 1 = 3
properties containing two characteristics each. So,
there are in fact only 20  3 = 17 individual
properties.
Using the Addition Rule for Probability
(cont.)
• The probability that the first condo Teresa
sees is either close to the subway or has a low
maintenance fee is 11 out of 17 approximately
64.71%.
Addition Rules for Probability
• Addition Rule for Probability
• For two events, E and F, the probability that E
or F occurs is given by the following formula.
P  E or F   P  E   P  F   P E and F 
Using the Addition Rule for Probability
(cont.)
P  close to subway or low fee
 P  close to subway   P low fee
P  close to subway and low fee
6 7 2 11
     0.6471
17 17 17 17
Using the Addition Rule for Probability
• Suppose that after a vote in the US Senate on a
proposed health care bill, the following table shows the
breakdown of the votes by party.
Votes on Health Care Bill
Democrat
Republican
Independent
Voted in Favor
Voted Against
23
43
2
21
7
4
• If a lobbyist stops a random senator after the vote,
what is the probability that this senator will either be a
Republican or have voted against the bill?
Using the Addition Rule for Probability
(cont.)
• Solution
There are a total of 100 US senators, all of
whom voted on this bill according to the table.
Of these senators, 43 + 7 = 50 are Republicans
and 21 + 7 + 4 = 32 voted against the bill.
However, 7 of the senators are both Republican
and voted against the bill.
Using the Addition Rule for Probability
(cont.)
• Thus, the Addition Rule would apply as follows.
P Republican or against 
 P Republican  P  against   P Republican and against
50 32
7
75 3




  0.75
100 100 100 100 4
• So the probability that the senator the lobbyist
stops will either be a Republican or have voted
against the bill is 75%.
Using the Addition Rule for Probability
Roll a pair of dice. What is the probability of
rolling either a total less than four or a total
equal to ten?
• Solution
P less than four or ten
 P less than four   P  ten  P less than four and ten
Using the Addition Rule for Probability
(cont.)
• We need to determine the number of
outcomes that give a total less than four and
the number of outcomes that give a total of
ten. Let’s list these outcomes in a table.
Totals Less Than Four
Totals of Ten
Using the Addition Rule for Probability
(cont.)
• By counting the outcomes, we see that there
are 3 outcomes that have totals less than four
and 3 outcomes that have totals of exactly
ten. Note that there are no outcomes that are
both less than four and exactly ten.
Using the Addition Rule for Probability
(cont.)
• Hence, we fill in the probabilities as follows.
P less than four or ten
 P less than four   P  ten  P less than four and ten
3 3 0
  
36 36 36
6

36
1
  0.1667
6
Addition Rules for Probability
• Addition Rule for Probability of Mutually
Exclusive Events
• If two events, E and F, are mutually exclusive,
then the probability that E or F occurs is given
by the following formula.
P  E or F   P  E   P  F 
Using the Addition Rule for Probability
of Mutually Exclusive Events
At a major exit on the interstate, past experience tells
us that the probabilities of a truck driver refueling at
each of the five possible gas stations are given in the
table below. Assuming that the truck driver will refuel
at only one of the stations (thus making the events
mutually exclusive), what
Probabilities of Refueling
is the probability that the
Gas Station
Probability
BP
0.0351
driver will refuel at Shell,
Chevron
0.1539
Exxon, or Chevron?
Exxon
0.2793
Shell
Texaco
0.3207
0.2110
Using the Addition Rule for Probability
of Mutually Exclusive Events (cont.)
• Solution
Since these three events are mutually exclusive,
we can add the individual probabilities together.
P  Shell or Exxon or Chevron
 P  Shell  P Exxon  P  Chevron
 0.3207  0.2793  0.1539
 0.7539
And more probability
Fundamental Counting Principle
• Bulloch Drug sells ice cream cones. They
have three different types of cones
(cake, waffle, and sugar) and 12
different types of ice creams.
• How many different combinations you
could make if you must choose one type
of cone and one type of ice cream?
Fundamental Counting Principle
• You could have:
• a sugar cone with each of the 12 types
of ice cream.
• a waffle cone with each of the 12 types
of ice cream.
• a cake cone with each of the 12 types of
ice cream.
3 x 12 = 36
Using the Fundamental Counting
Principle
The governing board of the local charity,
Mission Stateville, is electing a new vice
president and secretary to replace
outgoing board members. If the board
consists of 11 members who don’t already
hold an office, in how many different ways
can the two positions be filled if no one
may hold more than one office?
Using the Fundamental Counting
Principle
• Solution
There are two slots to fill in this example. Once
someone is chosen for one office, they cannot be
chosen for the other. The first slot may be filled with
any of the 11 board members. The second position has
one fewer to choose from, so there are only 10 choices
for it. Using the Fundamental Counting Principle, we
then multiply. There are
• 11 ⋅ 10 = 110 possible ways to elect the new officers.
Using the Fundamental Counting
Principle to Calculate Probability
• Robin prepares an afternoon snack for her
friends, Matthew and Lainey, every day. She
wants to give each friend one item. She has
the following snacks on hand: carrots, raisins,
crackers, grapes, apples, yogurt, and granola
bars. If she randomly chooses one snack for
Matthew and one snack for Lainey, what is the
probability that each friend gets the same
snack as yesterday?
Using the Fundamental Counting
Principle to Calculate Probability
• Solution
To begin, we need to count the number of ways in
which Robin can randomly choose a snack for her
friends. To do this, think of there being two slots to
fill—one for each friend. Because there is no
requirement that they have different snacks or the
same snack, there are 7 possibilities for each friend.
Therefore, there are 7 ⋅ 7 = 49 possible ways she
can prepare the snacks.
Using the Fundamental Counting
Principle to Calculate Probability
• Putting this information together, we can calculate
the probability.
P  choose same snack 
number of ways to choose same snack

total number of snacks
1
  0.0204
49
• Thus, there is about a 2% chance that each child will
eat the same thing two days in a row.
Probability of independent events
Two out of three people can roll their
tongue.
Nine out of ten people can wink.
What is the probability that and given
person can roll their tongue and wink?
Probability of independent events
• Solution:
• Because ones ability to roll their tongue
does not affect their ability to blink, the two
events are independent.
2 9
3
×
=
3 10 5
Multiplication Rules for Probability
Multiplication Rule for Probability
of Independent Events
• For two independent events E
and F, the probability that E and F
occur is given by the following
formula.
P  E and F   P  E   P  F 
Using the Multiplication Rule for
Probability of Independent Events
Choose two cards from a standard deck, with
replacement. What is the probability of choosing
a king and then a queen?
• Solution
Because the cards are replaced after each draw,
the probability of the second event occurring is
not affected by the outcome of the first event.
Thus, these two events are independent.
Using the Multiplication Rule for
Probability of Independent Events
• Using the Multiplication Rule for Probability of
Independent Events, we have the following.
P king and queen, with replacement 
 P king   P  queen
1
1
4 4
 
52 52
13
13
1

 0.0059
169
Using the Extended Multiplication Rule
for Probability of Independent Events
Assume that a study by Human Resources has
found that the probabilities of an employee
being written up for the following infractions are
the values shown in the following table. Assume
that each infraction is independent of the
others. What is the probability that a given
employee will be written up for being late for
work, taking unauthorized breaks, and leaving
early?
Using the Extended Multiplication Rule
for Probability of Independent Events
Probabilities of Being Written Up at Work
Infraction
Probability
Insubordination
Late for work
Failure to show up
for work
Leaving early
Taking unauthorized
breaks
0.1356
0.2478
0.1026
0.1954
0.3186
• Since these three events are independent, we can
apply the Multiplication Rule for Probability of
Independent Events to this situation.
Using the Extended Multiplication Rule
for Probability of Independent Events
 late for work and taking unauthorized 
P

breaks
and
leaving
early


 P late for work   P  taking unauthorized breaks 
 P leaving early 
 0.2478  0.3186  0.1954
 0.0154
Calculating Probability
What is the probability of drawing a king and
then a queen from a standard deck if the cards
are drawn without replacement?
Calculating Probability
What is the probability of drawing a king and then a
queen from a standard deck if the cards are drawn
without replacement?
• Solution
This situation is essentially the same as drawing two
cards at the same time from a standard deck. We
think of the experiment in two stages, though, for
ease in calculation. Begin by determining the
probability of drawing a king from a standard deck
of cards.
Calculating Probability
Since there are 4 kings in a deck of 52 cards, the
4 1
probability is
P king    .
52 13
Next, assume that when you drew the first card, it was
a king. What is the probability that you now draw a
queen? Well, since we are holding a king in our hand,
there are still 4 queens left in the deck of cards. But,
there are no longer 52 cards total, only 51. Thus,
4
P  queen, given king drawn first, without replacement  .
51
Calculating Probability
• We can now find the probability of the
multistage experiment by multiplying the two
individual probabilities.
P king and queen, without replacement 
 P king   P  queen, given king drawn first 
1
4 4 4
  
 0.0060
52 51 663
13
Calculating Conditional Probability
One card has already been chosen from a
standard deck without replacement. What is the
probability of now choosing a second card from
the deck and it being red, given that the first
card was a diamond?
Calculating Conditional Probability
One card has already been chosen from a standard
deck without replacement. What is the probability
of now choosing a second card from the deck and it
being red, given that the first card was a diamond?
• Solution
First, we must determine how many red cards are
left in the deck after the first pick. Because the first
card was a diamond (which is a red card), there are
only 25 red cards left in the deck instead of 26.
There are also only 51 cards total left in the deck.
Calculating Conditional Probability
(cont.)
• Thus, the conditional probability is calculated
as follows.
25
P red diamond 
51
 0.4902
Using the Multiplication Rule for
Probability of Dependent Events
What is the probability of choosing two face
cards in a row? Assume that the cards are
chosen without replacement.
Using the Multiplication Rule for
Probability of Dependent Events
• Solution
When the first card is picked, all 12 face cards
are available out of 52 cards. When the second
card is drawn, there are only 11 face cards left
out of the 51 cards remaining in the deck.
Using the Multiplication Rule for
Probability of Dependent Events
• Thus, we have the following.
P  face card and face card
 P  face card  P  face card face card
1
3
12 11
 
52 51
13
17
11
  0.0498
221
Using the Multiplication Rule for
Probability of Dependent Events
Assume that there are 17 men and 24 women in the
Rotary Club. Two members are chosen at random each
year to serve on the scholarship committee. What is
the probability of choosing two members at random
and the first being a man and the second being a
woman?
Using the Multiplication Rule for
Probability of Dependent Events
When the first member is picked, there are 17
men out of 41 members. When the second
member is picked, we assume that we have
already picked a man, so that leaves all 24
women, but only 40 remaining members.
Using the Multiplication Rule for
Probability of Dependent Events
• The calculation is as follows.
P man and woman  P man  P  woman man
3
17 24
 
41 40
5
51

205
 0.2488