MTH 1080: MATHEMATICAL MODES OF THOUGHT
PROBABILITY WORKSHEET PART II
Important Note: This is Part II of a series of two probability worksheets. This worksheet
provides a supplement to the exposition of probability theory provided by your instructor
and textbook (Pearson Custom Mathematics Textbook Ch. 2). These notes are not intended
to be a replacement for class notes or an alternative to reading the textbook.
3. Calculating Probabilities of Unions of Events
Unions of events are always at least as large as intersections of the same events. The union
of two events occurs if at least one of the two events occur. If we use A and B to symbolize
two events, then A ∪ B, read “A union B,” symbolizes the set of all outcomes that are either
in A or in B.
The union A ∪ B can be difficult to describe verbally, but the concept can be understood
quite clearly using our picture of the sample space S. If we consider the event A, described
as “doubles,” and the event B, described as “the sum of the pips is at least 9,” then the union
A ∪ B is simply all of the outcomes highlighted (in any color) in the picture below.
S={
p p
pp p
ppp p
pp pp p
pp p pp p
pp p
p pp , p
p p
p
, p p , p
p p
p
, pp p , pp
pp p
pp
, pp p , pp
ppp ,
ppp ,
ppp ,
ppp ,
p pp pp , p
p p pp pp , p p
p p p pp pp , p p p
pp pp pp pp , pp pp
pp p pp , p
pp p pp , p p
pp p pp , p p p
pp p pp , pp pp
pp
pp
pp p pp
p
, pp
ppp ,
ppp ,
pp p pp
p pp
pp p pp , pp p pp
pp p pp , pp p
pp
pp }
,
,
pp ,
pp ,
pp p pp
p pp
pp pp , pp p pp
pp pp , pp p
pp
pp
We can count all of the outcomes highlighted and see that it is 14. There are 36 possible
14
7
outcomes, so the probability of A ∪ B is P [A ∪ B] = 36
= 18
= 0.389. In other words, there
is 38.9% chance of either rolling doubles or rolling at least 9.
The colored visualization of the events in the sample space used for the previous example
allows us to discover an important probability rule. It is important to note that when we
qq qq
counted the number of outcomes in A ∪ B, we did not count the common outcomes of qqq qqq
qq qq
and qq qq qq qq twice, so the probability P [A ∪ B] is not typically the sum of P [A] and P [B]. This
causes trouble for students sometimes.
The probability P [A ∪ B] can still be calculated correctly using P [A], P [B] and P [A ∩ B]
using the following formula presented in the textbook called the general addition rule:
P [A ∪ B] = P [A] + P [B] − P [A ∩ B]
1
2
MTH 1080: MATHEMATICAL MODES OF THOUGHT PROBABILITY WORKSHEET PART II
The following calculation illustrates the general addition rule when used for the example
we already solved:
P [A ∪ B]
=
=
P
hn
p p p pp pp , pp pp
hn
p p , pp
+P
P
hn
−P
=
=
=
p p p pp pp , pp pp pp p pp , pp pp pp pp , pp p pp pp pp , pp p pp p p , p p p p p , p pp
pp pp
pp
. . . ppp ppp , p p p p , , p
hn
pp p pp ,
pp ,
pp pp pp pp ,
p p p p p p , pp pp
pp pp , pp pp pp p pp , . . .
oi
p , p p p p , p p p p p p , pp pp pp pp
pp p pp pp pp , pp p pp pp p pp , pp p pp pp pp , pp pp p p p , pp pp pp pp , pp pp pp p pp , pp pp pp pp oi
oi
pp pp , pp p pp pp p pp , pp pp p p
pp p pp pp p pp , pp pp pp p oi
P [A] + P [B] − P [A ∩ B]
10
6
2
+
−
36
36
36
14
= .389
36
Let us finish this section by considering the coin events defined in Section 2. We looked
at the experiment of flipping three fair coins, a dime, a nickel and a penny. We defined the
event E verbally as “all three coins have the same face,” and we defined the event F as “at
least two heads appear.” Highlighting both events gave us the following picture:
S={
HHH
, HHT , HT H , T HH ,
TTT
, T T H , T HT , HT T
}
There are five total outcomes highlighted, so the probability of E ∪ F occurring is P [A ∪ B] =
P [{HHH}]= 58 = .625. In other words, there is a 62.5% chance of either all faces being the
same or at least two heads appearing.
The following calculation illustrates the general addition rule when used for this example:
P [E ∪ F ]
hn
oi
HHH , HHT , HT H , T HH , T T T
oi
hn
oi
hn
oi
hn
HHH , HHT , HT H , T HH
+P
HHH , T T T
−P
HHH
= P
= P
= P [E] + P [F ] − P [E ∩ F ]
4 2 1
=
+ −
8 8 8
5
=
= .625
8
Work out solutions to the following exercises to help solidify this concept.
Exercies:
3.1. Suppose four coins, a quarter, a nickel, a penny and a dime are flipped. Let A be the
event that the quarter shows heads and B be the event that the dime shows heads. Calculate
the probability of the union A ∪ B.
3.2. Suppose we roll two fair, six-sided dice. Use the additional copies of the dice roll sample
space provided in Appendix A at the end of these notes to help visualize the solutions of the
following parts.
MTH 1080: MATHEMATICAL MODES OF THOUGHT
PROBABILITY WORKSHEET PART II
3
q
(1) Let A be the event that the first die shows qq and B be the event that the second die
q
shows qq. Compute the probability of the intersection A ∪ B.
(2) Suppose we roll two fair, six-sided dice. Let E be the event that at least one of the
q
qq
dice shows qq and F be the event that at least one of the dice shows qq qq. Compute the
probability of the intersection E ∪ F .
3.3. Let E1 be the event that first die has less than 4 pips and E2 be the event that the
second die has at least 4 pips. Compute the probability of the union E1 ∪ E2 .
3.4. Suppose four coins, a quarter, a dime, a nickel and a penny, are flipped. Let A be the
event that at least 3 tails appear and B be the event that the dime shows tails. Compute
the probability of the union A ∪ B.
4. Calculating Probabilities of Complement Events
This section is fast and quick to the point; our four dice/coin examples will use the same
four events A, B, E and F , that were used in the previous sections.
The complement of an event is just the nonoccurrence of the event. Given any event, E,
the event E must either occur or not occur; the event that it does not occur is called the
complement of E. We symbolize it with the notation E 0 .
The probability either E or E 0 occur is 1, because either the event occurs or it does not
so there is a 100% chance that one of these two things happen. However, E and E 0 have no
common outcomes, so we get the complement rule formulated as,
P [E] + P [E 0 ] = 1
This means that we can calculate the probability of E 0 by P [E 0 ] = 1 − P [E]. The four
examples below should clarify the concept and show the validity of the complement rule.
First consider the dice roll event “doubles” that we have defined to be A. The complement
event A0 includes all the outcomes in S that are not in A. Visually, these are the outcomes
highlighted below:
S={
p p
pp p
ppp p
pp pp p
pp p pp p
pp p p
p pp , p
p p
p
, p p , p
p p
p
, pp p , pp
pp p
pp
, pp p , pp
ppp , p
ppp , p p
ppp , ppp
p p p , pp pp
pp p
pp p
pp
, ppp p , ppp pp , ppp
pp p
p p
pp
, p p p , p p pp , p p
,
pp pp , p pp p pp , p
pp pp , p p pp p pp , p p
pp pp , p p p pp p pp , p p p
pp pp , pp pp pp p pp , pp pp
pp p
pp p
pp p
pp p
pp pp , pp p pp pp p pp , pp p pp pp p
pp pp , pp p pp p pp , pp p pp p }
We can see that there are exactly thirty outcomes in A0 by counting the outcomes highlighted
above. This implies that P [A0 ] = 30
= .833; in other words, there is a 83.3% chance of not
36
rolling doubles. Using the formula and the calculation of P [A] in Section 1, we can get the
same result using the formula,
P [A0 ] = 1 − P [A] = 1 − .167 = .833
4
MTH 1080: MATHEMATICAL MODES OF THOUGHT PROBABILITY WORKSHEET PART II
Next, considering the dice roll event B from above defined to be “the sum of the pips is
at least 9.” The outcomes that are not in B are highlighted below:
S={
p p
pp p
ppp p
pp pp p
pp p pp p
pp pp p
p pp , p
p p
p
, p p , p
p p
p
, pp p , pp
pp p
pp
, pp p , pp
ppp ,
ppp ,
ppp ,
ppp ,
p pp pp , p
p p pp pp , p p
p p p pp pp , p p p
pp pp pp pp , pp pp
pp p pp , p
pp p pp , p p
pp p pp , p p p
pp p pp , pp pp
pp p pp
pp
, pp
ppp ,
ppp ,
pp p pp
pp pp
pp p pp , pp p pp
pp p pp , pp pp
,
,
pp ,
pp ,
pp p pp
pp pp
pp pp , pp p pp
pp pp , pp pp
pp pp
pp pp
pp pp
pp pp
pp pp
pp pp }
26
= .722; alternatively, using the complement rule
Counting the outcomes gives us P [B 0 ] = 36
and our calculation of P [B] from Section 1, we can get that,
P [B 0 ] = 1 − P [B] = 1 − .278 = .722
In other words, there is a 72.2% chance of not rolling a sum of at least 9 pips.
Turning our attention to coin examples, we can consider the event E defined above as “all
three coins have the same face.” The outcomes that ar not in E are highlighted below:
S = { HHH , HHT , HT H , T HH ,
TTT
, T T H , T HT , HT T
}
Counting gives us that P [E 0 ] = 68 = .75; alternatively, the complement rule and our knowledge of P [E] gives us the same answer as
P [E 0 ] = 1 − P [E] = 1 − .25 = .75
In other words, there is a 75% chance of not flipping three of the same face.
Finally, defining F by “at least two heads appear,” we find that the occurrences that are
not in F can be highlighted as below:
S={
HHH
, HHT , HT H , T HH ,
TTT
, T T H , T HT , HT T
}
Counting gives us that P [E 0 ] = 84 = .50; alternatively, the complement rule and our knowledge of P [F ] gives us the same answer as
P [F 0 ] = 1 − P [F ] = 1 − .5 = .5
In other words, there is a 50% chance of not flipping at least two heads.
Exercies:
4.1. Suppose four coins, a quarter, a nickel, a penny and a dime are flipped. Compute the
probability of the complements of each of the following events.
(1) The event A defined to be “the quarter shows heads.”
(2) The event C defined to be “at least one heads appears.”
(3) The event F defined to be “all four coins show the same face.”
MTH 1080: MATHEMATICAL MODES OF THOUGHT
PROBABILITY WORKSHEET PART II
5
4.2. Suppose we roll two fair, six-sided dice. Compute the probability of the complements
of each of the following events.
q
(1) The event A defined to be “the first die shows qq ”
q
(2) The event C defined to be “at least one die shows qq ”
qqq
(3) The event F defined to be “both dice show ”
4.3. When rolling two fair, six-sided dice, what is the probability of not rolling a sum of 7
total pips?
4.4. Suppose four coins, a quarter, a dime, a nickel and a penny, are flipped. Let A be the
event that at least 3 tails appear and B be the event that the dime shows tails. Compute
the probability that neither A nor B occur; that is, compute P [(A ∪ B)0 ].
Appendix A: Six extra copies of the two-dice roll sample space
S={
q
q
qq q
qqq q
qq qq q
qqqqq q
qqq qqq q
S={
q
q
qq q
qqq q
qq qq q
qqqqq q
qqq qqq q
,
,
,
,
,
,
,
,
,
,
,
,
q qq
qq qq
qqq q q
qq qq q q
qqqqq q q
qqq qqq q q
q qq
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qqq q q
qq qq q q
qqqqq q q
qqq qqq q q
,
,
,
,
,
,
,
,
,
,
,
,
q qqq
q q qqq
qqq qqq
qq qq qqq
qqqqq qqq
qqq qqq qqq
q qqq
q q qqq
qqq qqq
qq qq qqq
qqqqq qqq
qqq qqq qqq
,
,
,
,
,
,
,
,
,
,
,
,
q qq qq
q q qq qq
qqq qq qq
qq qq qq qq
qqqqq
qqq qqq
qq qq
qq qq
q qq qq
q q qq qq
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qq qq qq qq
qqqqq
qqq qqq
qq qq
qq qq
,
,
,
,
,
,
,
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,
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,
q qqqqq
q q qqqqq
qqq qqqqq
qq qq qqqqq
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q q qqqqq
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qq qq qqqqq
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,
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q qqq qqq
q q qqq qqq
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}
q qqq qqq
q q qqq qqq
qqq qqq qqq
qq qq qqq qqq
qqqqq
qqq qqq
qqq qqq
qqq qqq
}
6
MTH 1080: MATHEMATICAL MODES OF THOUGHT PROBABILITY WORKSHEET PART II
Appendix A (continued): Six extra copies of the two-dice roll sample space
S={
q
q
qq q
qqq q
qq qq q
qqqqq q
qqq qqq q
S={
q
q
qq q
qqq q
qq qq q
qqqqq q
qqq qqq q
S={
q
q
qq q
qqq q
qq qq q
qqqqq q
qqq qqq q
S={
q
q
qq q
qqq q
qq qq q
qqqqq q
qqq qqq q
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
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,
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qq qq
qqq q q
qq qq q q
qqqqq q q
qqq qqq q q
q qq
qq qq
qqq q q
qq qq q q
qqqqq q q
qqq qqq q q
q qq
qq qq
qqq q q
qq qq q q
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qqq qqq q q
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qq qq q q
qqqqq q q
qqq qqq q q
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}
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}
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}