Complement
If A is an event, then the complement of A, written Ac, means
all the possible outcomes that are not in A.
For example, if A is the event “UNC wins at least 5 football
games”, then Ac is the event “UNC wins less than 5 football
games”.
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We can represent this by a Venn diagram, as follows:
S
A
c
A
Figure 1. Venn diagram for a single event A and its complement
Ac .
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The Law of Complementary Events states that
P (Ac) = 1 − P (A).
Example. In the game involving two throws of a die, if A is the
event “the total is 10 or greater”, then Ac is the event “the total
1 , so P (Ac ) = 5 .
is 9 or smaller”. We know P (A) = 6
6
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Disjoint events, Intersection and Union
Two events A and B are said to be disjoint if they cannot both
occur. This is represented by the following Venn diagram:
S
A
B
Figure 2. Venn diagram for two disjoint events A and B.
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For example, in two throws of a die, if A is the event “the total
is 10 or larger” and B is the event “the total is 3 or smaller”, it
is clear that A and B cannot both be true, so they are disjoint
events.
With any two events A and B, we define the intersection of A
and B, also written “A and B”, to be the event that A and B
both occur. With any two events A and B, we define the union
of A and B, also written “A or B”, to be the event that at least
one of A and B occurs.
Note that in common English, if we say “A or B”, that’s often
taken as excluding the possibility of both A and B occurring.
In the language of probability, “A or B” always includes the
possibility that both A and B might occur — unless they are
disjoint, in which case it is impossible.
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So another definition of disjoint events is: two events A and B
are disjoint if the intersection “A and B” is an impossible event.
The law of addition for disjoint events states that: If two events
A and B are disjoint, then
P (A or B) = P (A) + P (B).
Example. Consider the toss of two dice where A is the event
“the total is 10 or larger” and B is the event “the total is 3 or
1 and it is easy to
smaller”. We have already seen that P (A) = 6
1 . Therefore
see by similar reasoning that P (B) = 12
1
1
1
P (A or B) = P (A) + P (B) =
+
= .
6
12
4
We could also figure this out directly, by noting that the event “A
or B” consists of 9 outcomes of the sample space ((1,1),(1,2),(2,1)
9 = 1.
plus the 6 outcomes that comprise B) so the probability is 36
4
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The Law of Addition for Non-disjoint Events
S
A
B
Figure 3. Venn diagram for two disjoint events A and B.
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In this case the Law of Addition reads
P (A or B) = P (A) + P (B) − P (A and B).
Example. In a certain university, 52% of all students take a
statistics class, 23% take a computing course, and 7% take both.
What percentage of students take at least one of computing or
statistics?
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For a randomly chosen student let A be the event “the student
takes statistics”, and let B be the event “the student takes
computing”. The Venn diagram to represent this situation is:
S
A
B
Figure 4. Venn diagram for this problem.
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Applying the Law of Addition,
P (A or B) = 0.52 + 0.23 − 0.07 = 0.68.
In other words, 68% of students take at least one of Statistics
or Computing.
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Independent Events
Two events are said to be independent if the outcome of one
of them does not influence the other. For example, in sporting
events, the outcomes of different games are usually considered
independent even though that may not be true in a completely
strict and literal sense.
The multiplication rule for independent events says that if A and
B are independent,
P (A and B) = P (A) × P (B).
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Example: A football pundit states that the probability that UNC
will beat NC State is 0.4, while the probability that UNC will
beat Duke is 0.8. What is the probability that
1. UNC wins both games?
2. UNC wins at least one game?
3. UNC loses both games?
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Solution:
1. If A is the event “UNC beats State” and B is the event “UNC
beats Duke”, and if we assume these are independent events,
then the probability of A and B is 0.4 × 0.8 = 0.32.
2. Apply the Law of Addition:
P (A or B) = P (A)+P (B)−P (A and B) = 0.4+0.8−0.32 = 0.88.
3. Apply the Law of Complementary Events: “UNC loses both
games” is the complement of “UNC wins at least one game”,
so its probability is 1 − 0.88 = 0.12.
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Warning:
Don’t confuse the notions of “independent events” and “disjoint
events”. Independence means that the outcome of one event
does not influence the outcome of the other. Disjoint means
that if one event occurs then the other cannot occur — the very
opposite of independence!
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Conditional Probabilities
Consider the example (page 218 of text, referring to the Wimbledon tennis tournament),
A: “Federer misses his first serve”
B: “Federer misses his second serve”
We are told that Federer misses his first serve 36% of the time,
and that of all the times he misses his first serve, he also misses
his second serve 6% of the time.
What, then, is the probability he has a double fault?
Logically, the answer is 6% of 36%, or 0.06×0.36, which is about
0.02.
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Now let us rephrase this in the language of conditional probability.
We are told that the event A occurs 36% of the time, or in other
words P (A) = 0.36.
We are also told that, given that A has occurred, the event B
occurs 6% of the time. This is written in probability notation as
P (B | A) = 0.06.
The left hand side is read as the probability of B given A. In this
particular context, it would not make sense to talk about the
probability of B given Ac, though in other contexts, that would
make sense (e.g. free throws in basketball).
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The law of multiplication for conditional probabilities says
P (A and B) = P (A) × P (B | A).
Note that if we just interchange the role of A and B, we also get
P (A and B) = P (B) × P (A | B).
Finally, if A and B are independent, we get P (A | B) = P (A)
and P (B | A) = P (B) — that formalizes what is meant by saying
that the outcome of one event does not influence the outcome
of the other. But in that case, either of the last two formulas
reduces to
P (A and B) = P (A) × P (B)
as in our earlier formulation of the multiplication rule for independent events.
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Here is another (more complicated) example.
Consider the game in which a player tosses a die twice, and we
want to calculate the probability that the total of the two tosses
is at least 10. Define the events
A: The first throw is a 6.
B: The first throw is a 5.
C: The first throw is a 4.
D: The total of the two throws is at least 10.
Note that if the first throw is less than 4, it’s impossible for the
total to be 10 or higher. So
P (D) = P (A and D) + P (B and D) + P (C and D).
(1)
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1 . Given that A has occurred, D will occur if the
Now P (A) = 6
second throw produces any of 4, 5 or 6, and the probability of
1 . So we have
one of those outcomes is 3
or
6
2
1
1 1
1
1
×
=
.
P (A) = , P (D | A) = , P (A and D) =
6
2
6 2
12
Similarly
1
1
1 1
1
, P (D | B) = , P (B and D) =
×
=
,
6
3
6 3
18
1
1
1 1
1
P (C) =
, P (D | C) = , P (C and D) =
×
=
.
6
6
6 6
36
P (B) =
Therefore, (1) leads us to
1
1
1
1
P (D) =
+
+
=
12
18
36
6
giving the same answer as in our earlier calculation.
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