Math 1300: Section 8-1 Sample Spaces, Events, and Probability

Experiments
Sample Spaces and Events
Probability of an Event
Equally Likely Assumption

Math 1300 Finite Mathematics
Section 8-1: Sample Spaces, Events, and Probability

Jason Aubrey

Department of Mathematics
University of Missouri

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Jason Aubrey Math 1300 Finite Mathematics

Experiments

Probability theory is a branch of mathematics that has been
developed do deal with outcomes of random experiments. A
random experiment (or just experiment) is a situation
involving chance or probability that leads to results called
outcomes.



Experiments

Deﬁnition
The set S of all possible outcomes of an experiment a way
that in each trial of the experiment one and only one of the
outcomes (events) in the set will occur, we call the set S a
sample space for the experiment. Each element in S is
called a simple outcome, or simple event.



Experiments

Deﬁnition
An event E is deﬁned to be any subset of S (including the
empty set and the sample space S). Event E is a simple
event if it contains only one element and a compound
event if it contains more than one element.



Experiments

Deﬁnition
An event E is deﬁned to be any subset of S (including the
empty set and the sample space S). Event E is a simple
event if it contains only one element and a compound
event if it contains more than one element.
We say that an event E occurs if any of the simple events
in E occurs.


Experiments

Example: Suppose an experiment consists of simultaneously
rolling two fair six-sided dice (say, one red die and one green
die) and recording the values on each die.



Experiments

die) and recording the values on each die. Some possibilities
are (Red=1, Green=5)

(1, 5)



Experiments

are (Red=1, Green=5) or (Red=2, Green=2).

(1, 5) (2, 2)

What is the sample space S for this experiment?



Experiments

(1, 1) (1, 2) (1, 3) (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)



Experiments

To clarify, the sample space is always a set of objects. In this
case,
 
 (1, 1), (1, 2), (1, 3), (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),
 
S=
 (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)
 



Experiments

To clarify, the sample space is always a set of objects. In this
case,
 
 (1, 1), (1, 2), (1, 3), (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),
 
S=
 (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)
 

We can often use the counting techniques we learned in the
last chapter to determine the size of a sample space. In this
case, by the multiplication principle:

n(S) = 6 × 6 = 36


Experiments

Events are subsets of the sample space:



Experiments

A simple event is an event (subset) containing only one
outcome. For example,

E = {(3, 2)}

is a simple event.



Experiments

A simple event is an event (subset) containing only one
outcome. For example,

E = {(3, 2)}

is a simple event.
A compound event is an event (subset) containing more
than one outcome. For example,

E = {(3, 2), (4, 1), (5, 2)}

is a compound event.


Experiments

Events will often be described in words, and the ﬁrst step will
be to determine the correct subset of the sample space.



Experiments

be to determine the correct subset of the sample space. For
example
“A sum of 11 turns up” corresponds to the event
E = {(5, 6), (6, 5)}.
Notice that n(E) = 2.



Experiments

example
E = {(5, 6), (6, 5)}.
“The numbers on the two dice are equal” corresponds to
the event
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.
Here we have n(F ) = 6.



Experiments

example
E = {(5, 6), (6, 5)}.
“The numbers on the two dice are equal” corresponds to
the event
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.
Here we have n(F ) = 6.
“A sum less than or equal to 3” corresponds to the event:
G = {(1, 1), (1, 2), (2, 1)}
Here n(G) = 3

Experiments

Example: A wheel with 18 numbers on the perimeter is spun
and allowed to come to rest so that a pointer points within a
numbered sector.



Experiments

numbered sector.
(a) What is the sample space for this experiment?



Experiments

numbered sector.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}



Experiments

numbered sector.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

(b) Identify the event “the outcome is a number greater than
15”?



Experiments

numbered sector.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

15”?
E = {16, 17, 18}



Experiments

numbered sector.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

15”?
E = {16, 17, 18}
(c) Identify the event “the outcome is a number divisible by 12”?



Experiments

numbered sector.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

15”?
E = {16, 17, 18}
(c) Identify the event “the outcome is a number divisible by 12”?

E = {12}


Experiments

The ﬁrst step in deﬁning the probability of an event is to assign
probabilities to each of the outcomes (simple events) in the
sample space.



Experiments

sample space.
Suppose we ﬂip a fair coin twice. The sample space is

S = {HH, HT , TH, TT }



Experiments

sample space.
Suppose we ﬂip a fair coin twice. The sample space is


Since the coin is fair, each of the four outcomes is equally
likely, so P(HH) = P(HT ) = P(TH) = P(TT ) = 1 . 4



Experiments

Suppose the local meteorologist determines that the
chance of rain is 15%. As an experiment, we go out to
observe the weather. The sample space is

S = {Rain, No Rain}



Experiments


S = {Rain, No Rain}

The two outcomes here are not equally likely. We have
P(Rain) = 0.15 and P(No Rain) = 0.85.



Experiments


S = {Rain, No Rain}

The two outcomes here are not equally likely. We have
P(Rain) = 0.15 and P(No Rain) = 0.85.
Notice that in both cases, each probability was between zero
and one, and the sum of all of the probabilities was one.



Experiments

Deﬁnition (Probabilities of Simple Events)
Given a sample space

S = {e1 , e2 , . . . , en }

with n simple events, to each simple event ei we assign a real
number, denoted by P(ei ), called the probability of the event
ei .



Experiments


S = {e1 , e2 , . . . , en }

ei .
1 The probability of a simple event is a number between 0
and 1, inclusive. That is, 0 ≤ P(ei ) ≤ 1.



Experiments


S = {e1 , e2 , . . . , en }

ei .
1 The probability of a simple event is a number between 0
and 1, inclusive. That is, 0 ≤ P(ei ) ≤ 1.
2 The sum of the probabilities of all simple events in the
sample space is 1. That is,

P(e1 ) + P(e2 ) + · · · + P(en ) = 1.


Experiments

Two coin ﬂips. . . A possibly rainy day. . .

S = {Rain, No Rain}
e P(e)
e P(e) Rain 0.15
1
HH 4 No Rain 0.85
1
HT 4
1
TH 4
1
TT 4



Experiments

Example: Suppose a fair coin is ﬂipped twice. What is the
probability that exactly one head turns up.



Experiments

The event “exactly one head turns up” is the set

E = {HT , TH}



Experiments


E = {HT , TH}
1
We know that P(HT ) = 4 and P(TH) = 1 .
4



Experiments


E = {HT , TH}
1
We know that P(HT ) = 4 and P(TH) = 1 .
4

To determine P(E), just add the probabilities of the simple
events in E.
1 1 1
P(E) = P(HT ) + P(TH) = + =
4 4 2



Experiments

Deﬁnition (Probability of an Event E)
Given a probability assignment for the simple events in a
sample space S, we deﬁne the probability of an arbitrary
event E, denoted by P(E), as follows:



Experiments

1 If E is the empty set, then P(E) = 0.



Experiments

2 If E is a simple event, i.e. E = {ei }, then P(E) = P(ei ) as
deﬁned previously.



Experiments

3 If E is a compound event, then P(E) is the sum of the
probabilities of all the simple events in E.



Experiments

3 If E is a compound event, then P(E) is the sum of the
probabilities of all the simple events in E.
4 If E is the sample space S, then P(E) = P(S) = 1.



Experiments

Example: In a family with 3 children, excluding multiple births,
what is the probability of having exactly 2 girls? Assume that a
boy is as likely as a girl at each birth.



Experiments

First we determine the sample space S:

S = {GGG, GGB, GBG, BGG, GBB, BGB, BBG, BBB}



Experiments

First we determine the sample space S:

S = {GGG, GGB, GBG, BGG, GBB, BGB, BBG, BBB}

Since a boy is as likely as a girl at each birth, each of the 8
outcomes in S is equally likely; so each outcome has
1
probability 8 .



Experiments

Now we identify the event we wish to ﬁnd the probability of:

E = {GGB, GBG, BGG}



Experiments

Now we identify the event we wish to ﬁnd the probability of:

E = {GGB, GBG, BGG}

Therefore,

P(E) = P(GGB) + P(GBG) + P(BGG)
1 1 1 3
= + + =
8 8 8 8



Experiments

Procedure: Steps for Finding the Probability of an Event E



Experiments

1 Set up an appropriate sample space S for the experiment.



Experiments

2 Assign acceptable probabilities to the simple events in S.



Experiments

2 Assign acceptable probabilities to the simple events in S.
3 To obtain the probability of an arbitrary event E, add the
probabilities of the simple events in E.



Experiments

Recall two past examples. . .

Two coin ﬂips. . .


e P(e)
1
HH 4
1
HT 4
1
TH 4
1
TT 4



Experiments



S = {Rain, No Rain}
e P(e)
e P(e) Rain 0.15
1
HH 4 No Rain 0.85
1
HT 4
1
TH 4
1
TT 4



Experiments



S = {Rain, No Rain}
e P(e)
e P(e) Rain 0.15
1
HH 4 No Rain 0.85
1
HT 4
1 The outcomes are not equally
TH 4
1 likely.
TT 4

The outcomes are equally
likely. ../images/stackedlogo-bw-


Experiments

Sometimes we can assume that all outcomes in a sample
space are equally likely.



Experiments

Sometimes we can assume that all outcomes in a sample
space are equally likely.
If S = {e1 , e2 , . . . , en } is a sample space in which all
outcomes are equally likely, then we assign the probability
1
n to each outcome. That is

1
P(ei ) =
n
and we have. . .



Experiments

Theorem (Probability of an Arbitrary Event under an Equally
Likely Assumption)
If we assume that each simple event in a sample space S is as
likely to occur as any other, then the probability of an arbitrary
event E in S is given by

number of elements in E n(E)
P(E) = = .
number of elements in S n(S)



Experiments


(1, 5) (2, 2)



Experiments


(1, 5) (2, 2)

(a) What is the probability that the sum on the two dice comes
out to be 11?


Experiments

Firstly, since the dice are fair, for each die, the numbers 1-6
are all equally likely to turn up. So each possible pair of
numbers (1, 5), (3, 2), etc, is just as likely as any other. So,
we can make an equally likely assumption.



Experiments

We know from earlier that n(S) = 36.



Experiments

E = {(5, 6), (6, 5)}, so n(E) = 2.



Experiments

E = {(5, 6), (6, 5)}, so n(E) = 2.
Therefore
n(E) 2 1
P(E) = = =
n(S) 36 18



Experiments

(c) What is the probability that the numbers on the dice are
equal?



Experiments

equal?
The event here is
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}



Experiments

equal?
The event here is
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
So, n(F ) = 6



Experiments

equal?
The event here is
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
So, n(F ) = 6
Therefore,
n(F ) 6 1
P(F ) = = =
n(S) 36 6



Experiments

Example: 5 cards are drawn simultaneously from a standard
deck of 52 cards.
(a) Describe the sample space S. What is n(S)?



Experiments

deck of 52 cards.
Each outcome is a set of 5 cards chosen from the 52 available
cards. So, the sample space S can be described as

S = {all possible 5 card hands}



Experiments

deck of 52 cards.


How many 5-card hands can be drawn from a 52-card deck?



Experiments

deck of 52 cards.


From the previous chapter, we know this is

n(S) = C(52, 5) = 2, 598, 960



Experiments

deck of 52 cards.


From the previous chapter, we know this is

n(S) = C(52, 5) = 2, 598, 960

When the cards are dealt, each card is just as likely as any
other, so any ﬁve card hand is just as likely as any other. In
other words, we can make an equally likely assumption.

Experiments

(b)Find the probability that all of the cards are hearts.



Experiments

The event “all of the cards are hearts” is the set

E = {all 5 card hands with only hearts}



Experiments



Since there are 13 hearts in a standard deck of cards, we have

n(E) = C(13, 5) = 1287



Experiments



Since there are 13 hearts in a standard deck of cards, we have

n(E) = C(13, 5) = 1287

By the equally likely assumption

n(E) 1287
P(E) = = ≈ 0.000495
n(S) 2, 598, 960

or about 0.05%. ../images/stackedlogo-bw-


Experiments

(c) Find the probability that all the cards are face cards (that is,
jacks, queens or kings).



Experiments

The event “all the cards are face cards” is the set

F = {all 5 card hands consisting only of face cards}



Experiments



There are a total of 4 × 3 = 12 face cards. So,

n(F ) = C(12, 5) = 792



Experiments



There are a total of 4 × 3 = 12 face cards. So,

n(F ) = C(12, 5) = 792

By the equally likely assumption

n(F ) 792
P(F ) = = ≈ 0.000305
n(S) 2, 598, 960
or about 0.03%.

Experiments

(d) Find the probability that all the cards are even. (Consider
aces to be 1, jacks to be 11, queens to be 12 and kings to be
13).



Experiments

13).
The event “all the cards are even is the set

G = {all 5 card hands consisting of only even cards}



Experiments

13).


There are 6 even cards per suit (2,4,6,8,10,Q); so there are a
total of 20 even cards in a deck. So,

n(G) = C(20, 6) = 38, 760.



Experiments

13).


There are 6 even cards per suit (2,4,6,8,10,Q); so there are a
total of 20 even cards in a deck. So,

n(G) = C(20, 6) = 38, 760.

By the qually likely assumption,
n(G) 38, 760
P(G) = = ≈ 0.0149
n(S) 2, 598, 960

or about 14.9%.

Math 1300: Section 8-1 Sample Spaces, Events, and Probability

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