Union, Intersection, Complement and Odds in Probability

Union and Intersection
Complement of an Event
Odds
Applications to Empirical Probability

Math 1300 Finite Mathematics
Section 8-2: Union, Intersection, and Complement of Events;
Odds

Jason Aubrey

Department of Mathematics
University of Missouri

../images/stackedlogo-bw-

Jason Aubrey Math 1300 Finite Mathematics

Odds

In this section, we will develop the rules of probability for
compound events (more than one simple event) and will
discuss probabilities involving the union of events as well as
intersection of two events.



Odds

If A and B are two events in a sample space S, then the union
of A and B, denoted by A ∪ B, and the intersection of A and B,
denoted by A ∩ B, are deﬁned as follows:
Deﬁnition (Union: A ∪ B)
A ∪ B = {e ∈ S|e ∈ A or e ∈ B}

A B



Odds

If A and B are two events in a sample space S, then the union
of A and B, denoted by A ∪ B, and the intersection of A and B,
denoted by A ∩ B, are defined as follows:
Definition (Union: A ∪ B)
A ∪ B = {e ∈ S|e ∈ A or e ∈ B}

A B

The event “A or B” is defined as the set A ∪ B.

Odds

Deﬁnition (Intersection: A ∩ B)

A ∩ B = {e ∈ S|e ∈ A and e ∈ B}

A B



Odds

Deﬁnition (Intersection: A ∩ B)

A ∩ B = {e ∈ S|e ∈ A and e ∈ B}

A B

The event “A and B” is deﬁned as the set A ∩ B. ../images/stackedlogo-bw-


Odds

Example: Consider the sample space of equally likely events
for the rolling of a single fair die: S = {1, 2, 3, 4, 5, 6}



Odds

(a) What is the probability of rolling an odd number and a prime
number?



Odds

number?
Let A be the event “an odd number is rolled”. Let B be the event
“a prime number is rolled”.



Odds

number?
Since only the outcomes 3 and 5 are both odd and prime,
A ∩ B = {3, 5}.



Odds

number?
Since only the outcomes 3 and 5 are both odd and prime,
A ∩ B = {3, 5}.
By the equally likely assumption,

n(A ∩ B) 2 1
P(A ∩ B) = = =
n(S) 6 3


Odds

(b) What is the probability of rolling an odd number or a prime
number?



Odds

number?
Again, A = {1, 3, 5} and B = {2, 3, 5} so



Odds

number?
Again, A = {1, 3, 5} and B = {2, 3, 5} so

A ∪ B = {1, 2, 3, 5}



Odds

number?
Again, A = {1, 3, 5} and B = {2, 3, 5} so

A ∪ B = {1, 2, 3, 5}

By the equally likely assumption

n(A ∪ B) 4 2
P(A ∪ B) = = =
n(S) 6 3



Odds

Suppose that an event E is

E = A ∪ B.



Odds


E = A ∪ B.

Is P(E) = P(A) + P(B)?



Odds


E = A ∪ B.

Is P(E) = P(A) + P(B)?
Only if A and B are mutually exclusive (disjoint), that is,
if A ∩ B = ∅.



Odds


E = A ∪ B.

Is P(E) = P(A) + P(B)?
Only if A and B are mutually exclusive (disjoint), that is,
if A ∩ B = ∅.
In this case, P(A ∪ B) is the sum of the probabilities of all
of the simple events in A plus the sum of the probabilities
of all of the simple events in B.



Odds

But what happens if A and B are not mutually exclusive;
that is, what if A ∩ B = ∅?



Odds

But what happens if A and B are not mutually exclusive;
that is, what if A ∩ B = ∅?
In this case, we must use a version of the addition principle
for probability.



Odds

Deﬁnition (Probability of a Union of Two Events)
For any events A and B,

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

If A and B are mutually exclusive, then

P(A ∪ B) = P(A) + P(B)



Odds

Example Let E and F be events in a sample space S. If
P(E) = 0.35, P(F ) = 0.25 and P(E ∪ F ) = 0.55 then what is
P(E ∩ F )?



Odds

P(E ∩ F )?
Notice here that P(E ∪ F ) = P(E) + P(F ), so we know that the
events E and F are not mutually exclusive; that is,
P(E ∩ F ) = 0.



Odds

P(E ∩ F )?
P(E ∩ F ) = 0.
To ﬁnd P(E ∩ F ) we use the addition principle:

P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F )



Odds

P(E ∩ F )?
P(E ∩ F ) = 0.

P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F )
0.55 = 0.35 + 0.25 − P(E ∩ F )



Odds

P(E ∩ F )?
P(E ∩ F ) = 0.

P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F )
0.55 = 0.35 + 0.25 − P(E ∩ F )
P(E ∩ F ) = 0.35 + 0.25 − 0.55 = 0.05



Odds

Example: A single card is drawn from a deck of cards. Find the
probability that the card is a jack or club.



Odds

Let E = “the set of jacks”, and F = “the set of clubs”. Are E
and F mutually exclusive?



Odds

and F mutually exclusive? No because a card can be both a
jack and a club?



Odds

jack and a club?

Now, n(E) = 4 - there are four jacks; by the equally likely
assumption,
n(E) 4
P(E) = = .
n(S) 52



Odds

jack and a club?

Now, n(E) = 4 - there are four jacks; by the equally likely
assumption,
n(E) 4
P(E) = = .
n(S) 52
Also, n(F ) = 13 - there are thirteen clubs; by the equally
likely assumption,
n(F ) 13
P(F ) = = .
n(S) 52 ../images/stackedlogo-bw-


Odds

Finally, n(E ∩ F ) = 1 - there is one jack that is also a club;
by the equally likely assumption,

1
P(E ∩ F ) = .
52
Now we can use the addition principle to get



Odds


1
P(E ∩ F ) = .
52

P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F )



Odds


1
P(E ∩ F ) = .
52

P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F )
4 13 1
P(E ∪ F ) = + −
52 52 52



Odds


1
P(E ∩ F ) = .
52

P(E ∪ F ) = P(E) + P(F ) − P(E ∩ F )
4 13 1
P(E ∪ F ) = + −
52 52 52
16 4
= =
52 13


Odds

Example: Three coins are tossed. Assume they are fair coins.
(Tossing three coins is the same experiment as tossing one
coin three times.)



Odds

coin three times.)
(a) Use the multiplication principle to calculate the total number
of outcomes in the sample space.



Odds

coin three times.)

n(S) = (2)(2)(2) = 8



Odds

coin three times.)

n(S) = (2)(2)(2) = 8

(b) Find the probability of ﬂipping at least two tails.



Odds

coin three times.)

n(S) = (2)(2)(2) = 8

Let E be the event of ﬂipping at least two tails.



Odds

coin three times.)

n(S) = (2)(2)(2) = 8

Let A be the event that exactly two tails are ﬂipped.



Odds

coin three times.)

n(S) = (2)(2)(2) = 8

Let A be the event that exactly two tails are ﬂipped.
Let B be the event that exactly three tails are ﬂipped.


Odds

Notice that E = A ∪ B and that A and B are mutually
exclusive; that is,

E = A ∪ B and A ∩ B = ∅



Odds

exclusive; that is,

E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)



Odds

exclusive; that is,

E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)
n({HTT , THT , TTH}) 3
P(A) = =
n(S) 8



Odds

exclusive; that is,

E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)
P(A) = =
n(S) 8
n({TTT }) 1
P(B) = =
n(S) 8



Odds

exclusive; that is,

E = A ∪ B and A ∩ B = ∅

So P(E) = P(A ∪ B) = P(A) + P(B)
P(A) = =
n(S) 8
n({TTT }) 1
P(B) = =
n(S) 8
Therefore,
3 1 1
P(E) = P(A) + P(B) = + =
8 8 2


Odds

Suppose that we divide a ﬁnite sample space

S = {e1 , . . . , en }

into two subsets E and E such that

E ∩ E = ∅.



Odds


S = {e1 , . . . , en }


E ∩ E = ∅.

That is, E and E are mutually exclusive, and

E ∪ E = S.



Odds


S = {e1 , . . . , en }


E ∩ E = ∅.


E ∪ E = S.

Then E is called the complement of E relative to S.



Odds


S = {e1 , . . . , en }


E ∩ E = ∅.


E ∪ E = S.

Then E is called the complement of E relative to S. The set
E contains all the elements of S that are not in E.


Odds

Furthermore,

P(S) = P(E ∪ E )
= P(E) + P(E ) = 1



Odds

Furthermore,

P(S) = P(E ∪ E )
= P(E) + P(E ) = 1

Therefore,

P(E) = 1 − P(E ) P(E ) = 1 − P(E)



Odds

Furthermore,

P(S) = P(E ∪ E )
= P(E) + P(E ) = 1

Therefore,

P(E) = 1 − P(E ) P(E ) = 1 − P(E)

Many times it is easier to ﬁrst compute the probability that and
event won’t occur, and then use that to ﬁnd the probability that
the event will occur.



Odds

Example: If A and B are mutually exclusive events with
P(A) = 0.6 and P(B) = 0.3, ﬁnd the following probabilities:



Odds


(a) P(A ∩ B)



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0
(b) P(A ∩ B)



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0
(b) P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0 = 0.9



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0
(b) P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0 = 0.9
(c) P(A )



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0
(b) P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0 = 0.9
(c) P(A )
P(A ) = 1 − P(A) = 1 − 0.6 = 0.4



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0
(b) P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0 = 0.9
(c) P(A )
P(A ) = 1 − P(A) = 1 − 0.6 = 0.4
(d) P(A ∩ B )



Odds


(a) P(A ∩ B)
P(A ∩ B) = 0
(b) P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0 = 0.9
(c) P(A )
P(A ) = 1 − P(A) = 1 − 0.6 = 0.4
(d) P(A ∩ B )
Since A ∩ B = ∅, A ⊆ B so A ∩ B = A and
P(A ∩ B ) = P(A) = 0.6


Odds

Example: What is the probability that when two dice are
tossed, the number of points on each die will not be the same?



Odds

This is the same as saying that doubles will not occur. For
example,



Odds

example,

E be the set of all rolls of two dice which do not result in
doubles. Mathematically we can represent this as
E = {(n, m)|1 ≤ n, m ≤ 6 and n = m}
We wish to ﬁnd P(E).



Odds

example,

E be the set of all rolls of two dice which do not result in
doubles. Mathematically we can represent this as
E = {(n, m)|1 ≤ n, m ≤ 6 and n = m}
We wish to ﬁnd P(E). Let S be be the sample space for this
experiment.
S = {(n, m)|1 ≤ n, m ≤ 6} and n(S) = 36 ../images/stackedlogo-bw-


Odds

Here we have



Odds

Here we have

E = {(n, m)|1 ≤ n, m ≤ 6 and n = m}



Odds

Here we have

E = {(n, m)|1 ≤ n, m ≤ 6 and n = m}
= {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}



Odds

Here we have

E = {(n, m)|1 ≤ n, m ≤ 6 and n = m}
= {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Since
n(E ) 6 1
P(E ) = = =
n(S) 36 6



Odds

Here we have

E = {(n, m)|1 ≤ n, m ≤ 6 and n = m}
= {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Since
n(E ) 6 1
P(E ) = = =
n(S) 36 6
we have
1 5
P(E) = 1 − P(E ) = 1 − =
6 6



Odds

Example: A coin is tossed 5 times. What is the probability that
heads turn up at least once?



Odds

Let E represent the event “heads turns up at least once” and let
S represent the sample space. Notice ﬁrst that

S = {HTTHT , HHHTT , THHHT , HTHTH, TTTTT , HHHHT , . . .}



Odds



We assume the coin is fair so that we may also assume that all
of the outcomes in the sample space are equally likely. What is
n(S)?



Odds



n(S)?
n(S) = (2)(2)(2)(2)(2) = 32



Odds



n(S)?
n(S) = (2)(2)(2)(2)(2) = 32
E contains all outcomes that have at least one H. E.g. HTTHT ,
HHHTT , etc.


Odds

What is in the set E ?



Odds

What is in the set E ? The opposite of “heads turn up at least
once” is “heads do not turn up at all.” So,

1
E = {TTTTT } and P(E ) =
32



Odds


1
32
Therefore,
1 31
P(E) = 1 − P(E ) = 1 − =
32 32



Odds


1
32
Therefore,
1 31
P(E) = 1 − P(E ) = 1 − =
32 32
Tip: Consider using complements whenever you encounter a
probability (or even counting problems) that contains the phrase
“at least once”.


Odds

Example: A shipment of 40 precision parts, including 8 that are
defective, is sent to an assembly plant. The quality control
division selects 10 at random for testing and rejects the
shipment if 1 or more in the sample are found defective. What
is the probability that the shipment will be rejected?



Odds

Example: A shipment of 40 precision parts, including 8 that are
defective, is sent to an assembly plant. The quality control
division selects 10 at random for testing and rejects the
shipment if 1 or more in the sample are found defective. What
is the probability that the shipment will be rejected?

Notice ﬁrst that the question
What is the probability that the shipment will be
rejected?

is really asking
What is the probability that the 10 parts selected for
testing contain at least one defective part?


Odds

Let S be all possible selections of 10 parts from the shipment of
40.



Odds

40.
n(S) = C(40, 10) = 847, 660, 528



Odds

40.
n(S) = C(40, 10) = 847, 660, 528
Let E be the set of all selections of 10 parts that contain at least
one defective part. We want to ﬁnd P(E).



Odds

40.
n(S) = C(40, 10) = 847, 660, 528
Notice that every set of 10 parts either



Odds

40.
n(S) = C(40, 10) = 847, 660, 528
contains at least one defective part (so is in E), or



Odds

40.
n(S) = C(40, 10) = 847, 660, 528
contains no defective parts



Odds

40.
n(S) = C(40, 10) = 847, 660, 528
contains no defective parts
Thus E is the set of all selections of 10 parts that contain no
defective parts.



Odds

The shipment of 40 parts contains 8 that are defective. To pick
10 that have no defective parts, we choose from the 32 that are
not defective, so



Odds

not defective, so

n(E ) = C(32, 10) = 64, 512, 240



Odds

not defective, so

n(E ) = C(32, 10) = 64, 512, 240

Therefore,

n(E )
P(E) = 1 − P(E ) = 1 −
n(S)



Odds

not defective, so

n(E ) = C(32, 10) = 64, 512, 240

Therefore,

n(E )
P(E) = 1 − P(E ) = 1 −
n(S)
64, 512, 240
=1− ≈ 1 − 0.0761
847, 660, 528



Odds

not defective, so

n(E ) = C(32, 10) = 64, 512, 240

Therefore,

n(E )
P(E) = 1 − P(E ) = 1 −
n(S)
64, 512, 240
=1− ≈ 1 − 0.0761
847, 660, 528
≈ 0.9239



Odds

not defective, so

n(E ) = C(32, 10) = 64, 512, 240

Therefore,

n(E )
P(E) = 1 − P(E ) = 1 −
n(S)
64, 512, 240
=1− ≈ 1 − 0.0761
847, 660, 528
≈ 0.9239

So there is about a 92.4% chance that the shipment will be
rejected.

Odds



Odds

Deﬁnition (From Probabilities to Odds)
If P(E) is the probability of the event E, then
1 the odds for E are given by

P(E) P(E)
= , P(E) = 1.
1 − P(E) P(E )



Odds


P(E) P(E)
= , P(E) = 1.
1 − P(E) P(E )

2 the odds against E are given by

1 − P(E) P(E )
= , P(E) = 0
P(E) P(E)



Odds


P(E) P(E)
= , P(E) = 1.
1 − P(E) P(E )

2 the odds against E are given by

1 − P(E) P(E )
= , P(E) = 0
P(E) P(E)

Note: When possible, odds are to be expressed as ratios of
whole numbers. ../images/stackedlogo-bw-


Odds

Example: Given the following probabilities for an event E, ﬁnd
the odds for and against E:



Odds

3
P(E) = 5



Odds

P(E) = 35
Since P(E) = 3/5 we know P(E ) = 1 − P(E) = 2/5.



Odds

P(E) = 35
Then the odds for E are
P(E) 3/5 3
= =
P(E ) 2/5 2



Odds

P(E) = 35
P(E) 3/5 3
= =
P(E ) 2/5 2

And the odds against E are

P(E ) 2/5 2
= =
P(E) 3/5 3



Odds

P(E) = 0.35



Odds

P(E) = 0.35
Since P(E) = 0.35 we know P(E ) = 1 − 0.35 = 0.65



Odds

P(E) = 0.35
P(E) 0.35 7
= =
P(E ) 0.65 13



Odds

P(E) = 0.35
P(E) 0.35 7
= =
P(E ) 0.65 13

And the odds against E are

P(E ) 0.65 13
= =
P(E) 0.35 7



Odds

Example: Find the odds in favor of rolling a total of seven when
two dice are tossed.



Odds

Let E be the event that the sum of the two dice is seven. So,

E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}



Odds


E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

n(E) = 6, n(S) = 36,



Odds


E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

n(E) = 6, n(S) = 36,
6 30
P(E) = and P(E ) =
36 36



Odds


E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

n(E) = 6, n(S) = 36,
6 30
P(E) = and P(E ) =
36 36
Therefore
P(E) 6/36 6 1
= = =
P(E ) 30/36 30 5


Odds

a
If the odds for an event E are , then the probability of E
b
is,
a
P(E) =
a+b



Odds

a
If the odds for an event E are , then the probability of E
b
is,
a
P(E) =
a+b
a
If the odds against an event E are then the probability of
b
E is
b
P(E) =
a+b



Odds

Example: If in repeated rolls of two fair dice the odds against
rolling a 6 before rolling a 7 are 6:5, what is the probability of
rolling a 6 before rolling a 7?



Odds

Let E be the event “a 6 is rolled before a 7 is rolled”.



Odds

odds against E are 6:5



Odds

odds against E are 6:5
Therefore,
5 5
P(E) = =
6+5 11



Odds

Example: The data below was obtained from a random survey
of 1,000 residents of a state. The participants were asked their
political afﬁliations and their preferences in an upcoming
gubernatorial election (D = Democrat, R = Republican, U =
Unafﬁliated. )

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
No Preference 50 20 15 85
Totals 500 350 150 1,000



Odds

Example: The data below was obtained from a random survey
of 1,000 residents of a state. The participants were asked their
political affiliations and their preferences in an upcoming
gubernatorial election (D = Democrat, R = Republican, U =
Unaffiliated. )

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
Totals 500 350 150 1,000

If a resident of the state is selected at random, what is the
empirical probability that the resident is not affiliated with a
political party or has no preference? What are the odds for this
event?

Odds

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
Totals 500 350 150 1,000
We are looking for P(U ∪ N):



Odds

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
Totals 500 350 150 1,000
P(U ∪ N) = P(U) + P(N) − P(U ∩ N)



Odds

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
Totals 500 350 150 1,000
P(U ∪ N) = P(U) + P(N) − P(U ∩ N)
150 85 15
= + −
1000 1000 1000



Odds

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
Totals 500 350 150 1,000
P(U ∪ N) = P(U) + P(N) − P(U ∩ N)
150 85 15
= + −
1000 1000 1000
220
= = 0.22 or 22%
1000



Odds

D R U Totals
Candidate A 200 100 85 385
Candidate B 250 230 50 530
Totals 500 350 150 1,000
P(U ∪ N) = P(U) + P(N) − P(U ∩ N)
150 85 15
= + −
1000 1000 1000
220
= = 0.22 or 22%
1000
Then the odds for this event are
22 22 11
= =
100 − 22 78 39 ../images/stackedlogo-bw-


Union, Intersection, Complement and Odds in Probability

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