STATS 120A/281A
Lecture 2
Sevan Koko Gulesserian
University of California, Irvine
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Probability Theory
Disjoint/mutually exclusive events.
Say we have two sets/events, A and B.
We say that A and B are mutually exclusive or disjoint
sets/events if A ∩ B = ∅.
This is to say the sets A and B have no overlap (nothing in
common, no overlap in a Venn diagram). They are disjoint
sets.
With respect to events, this is saying that events A and B
cannot occur together. If A occurs, then B cannot occur, and
vice versa.
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Probability Theory
Disjoint/mutually exclusive events.
Say we have two sets/events, A and B.
Example: Roll a single die. Let A be the event that an even
number is rolled and B be the event that an odd number is
rolled.
Then A = {2, 4, 6} and B = {1, 3, 5}. Therefore A ∩ B = ∅.
Therefore say that A and B are disjoint or mutually exclusive.
Another example: Roll a pair of dice and let A={both dice are
odd} and B={sum of the dice is equal to 12}.
Events A and B are mutually exclusive (they cannot both
occur, only way to get a sum of 12 is to roll both 6’s).
We will develop a probability based result to see if events are
mutually exclusive/disjoint.
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Probability Theory
Probability.
Will view probability as a function.
Denoted as Pr(.) or P(.).
It is a function that will map the set of all events to the
interval [0,1].
That is to say, it will assign a probability to an event between
0 and 1. 0 meaning it cannot occur and 1 meaning it will
definitely occur.
Commonly convert probabilities to percentages. For example
0.5 is 50% ,1 is 100%, and 0.475 is 47.5%.
Think of probability as areas of the circles within a Venn
diagram (where the rectangle that holds the circles has an area
of 1).
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Probability Theory
The axioms of probability.
Let S be the sample space of some experiment, and A be a specific
event.
1. 0 ≤ Pr (A) ≤ 1 for all events A.
That is to say probability of an event cannot be negative or
above 1. Must be between 0 and 1.
2. Pr (S) = 1.
The probability of the sample space occurring is 1. That is to
say an element of the sample space must occur when
conducting the experiment.
3. For any infinite sequence of disjoint events A1 , A2 , ..., we
∞
∞
P
have: Pr ( ∪ Ai ) =
Pr (Ai ).
i=1
i=1
The probability of the infinite union of these disjoint sets is the
sum of their probabilities.
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Set Theory Basics
Now say we have sets A1 , A2 , A3 , ... (or can think of sets
A, B, C , ...).
We say that sets (or events) are mutually exclusive/disjoint if
the intersection between any of these two sets is the null set.
Notationally Ai ∩ Aj = ∅ for all i 6= j.
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Probability Theory
Probability of the null set.
A theorem that results from the axioms listed is that
Pr (∅) = 0.
To prove this, let Ai = ∅ for i=1,2,3,....
The Ai ’s create a sequence of disjoint events since ∅ ∩ ∅ = ∅.
∞
∞
i=1
i=1
Also, ∪ Ai = ∪ ∅ = ∅.
Using the 3rd axiom, we have that
∞
∞
∞
P
P
Pr ( ∪ Ai ) = Pr (∅) =
Pr (Ai ) =
Pr (∅).
i=1
i=1
i=1
This implies that Pr (∅) = 0 since Pr (∅) =
∞
P
Pr (∅).
i=1
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Probability Theory
With just the axioms of probability, a few more theorems can be
established.
Let A and B be events within the sample space S.
Theorem: For all finite sequences of disjoint events
n
n
P
Pr (Ai ).
A1 , A2 , ..., An , we have Pr ( ∪ Ai ) =
i=1
Theorem:
P(Ac )
i=1
= 1 − P(A).
Theorem: If A is a subset of B (that is to say A ⊆ B) then
P(A) ≤ P(B).
Theorem: P(A ∩ B c ) = P(A) − P(A ∩ B).
This can be rearranged to get P(A) = P(A ∩ B) + P(A ∩ B c ).
Theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
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Probability Theory
To prove P(Ac ) = 1 − P(A):
The area of S is 1 (since P(S)=1), thus P(A) + P(Ac ) = 1.
More formally, we have that P(S) = P(A ∪ Ac ) where A and Ac
are disjoint. Thus, P(S) = 1 = P(A) + P(Ac ).
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Probability Theory
To prove that if A is a subset of B (that is to say A ⊆ B) then
P(A) ≤ P(B):
Since A is contained in B, then P(A) ≤ P(B).
More formally, B = A ∪ (Ac ∩ B) (since A ⊆ B) where A and
Ac ∩ B are disjoint.
Thus, P(B) = P(A) + P(Ac ∩ B) which implies P(B) ≥ P(A)
since P(Ac ∩ B) ≥ 0.
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Probability Theory
To prove P(A ∩ B c ) = P(A) − P(A ∩ B) :
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Probability Theory
To prove P(A ∪ B) = P(A) + P(B) − P(A ∩ B):
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Probability Theory
Stated that if A and B are mutually exclusive events, then
both A and B cannot occur together.
That is to say A ∩ B = ∅.
The probability of such an event is 0.
Since P(A ∩ B) = P(∅) = 0.
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Probability Theory
Following from theorem P(A ∪ B) = P(A) + P(B) − P(A ∩ B), if
A and B are disjoint/mutually exclusive events then the formula
simplifies to:
P(A ∪ B) = P(A) + P(B) , since P(A ∩ B) = 0.
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Probability Theory
Example: A student will get into business graduate school if both
events A and B occur. Event A is the they score above 600 on the
GMAT entrance exam and event B is that they obtain a GPA above
3.5 in undergraduate work.
A student estimates that P(A)=0.62 and P(B)=0.81.
They also estimate that there is a 15% chance that neither A nor B
will occur.
What is the probability that they get into business graduate school.
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Probability Theory
Example:
What we have is that P(A)=0.62, P(B)=0.81 and
P(Ac ∩ B c ) = 0.15.
Want to find P(A ∩ B), which means both A and B occur,
which then means the student gets into graduate school.
Note that by DeMorgan’s law that Ac ∩ B c = (A ∪ B)c .
We also know that P(Ac ) = 1 − P(A).
And we also know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
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Probability Theory
Thus:
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
= P(A) + P(B) − [1 − P((A ∪ B)c )]
= P(A) + P(B) − [1 − P(Ac ∩ B c )]
= 0.62 + 0.81 − (1 − 0.15)
= 0.58
Now say you want to find out P(A ∪ B) (the probability that A or
B occurs).
We have that
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.62 + 0.81 − 0.58 = 0.85.
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Probability Theory
Equally likely outcomes:
Let S = {s1 , s2 , ..., sn } with pi = P(si ) for i=1,2,...,n.
From the axioms, we know that 0 ≤ pi ≤ 1 and that
n
P
pi = 1.
P(S) =
i=1
Now if pi =
1
n
for all i, we call S a simple space.
We have equally likely outcomes (each outcome is equally
likely of occurring).
For any event A within S (A ⊂ S), we have P(A) =
m is the number of outcomes in the event A.
m
n
where
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Probability Theory
Example: Roll a die once. The sample space is S = {1, 2, 3, 4, 5, 6},
and each outcome is equally likely. Thus each element has
probability of 61 of occurring.
Note that P(S) =
6
P
1
i=1
6
= 1.
Let event A be the event of rolling either a 1 or a 6.
Event A has two possible outcomes (1 or 6).
Thus, P(A) = 62 .
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Probability Theory
Example: Roll a pair of dice once and record the maximum number
that shows (so if you roll a 2 and a 4 then you record 4, if you roll a
4 and a 4, the maximum is 4).
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Probability Theory
The sample space of the maximum of the two rolls is
S = {1, 2, 3, 4, 5, 6}.
Each element of the sample space is not equally likely, but
each of the 36 different combos of rolls is equally likely.
Say event A is that the maximum is a 1. There is only 1 way
1
this can happen (roll a 1 on each die). Thus P(A) = 36
.
Now say the event A is that the maximum is a 2. There are 3
ways this can occur, roll a (2,2) or (1,2) or (2,1). Thus
3
P(A) = 36
.
Can go on in this manner and see that the probability of a
5
7
maximum of 3 is 36
, of a maximum of 4 is 36
, of a maximum
9
11
of 5 is 36 and of a maximum of 6 is 36 .
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