Download Math 425 Introduction to Probability Lecture 5

The probability model
The probability model
Math 425
Introduction to Probability
Lecture 5
+ The basis of probability is the experiment:
An experiment is a repeatable procedure that has a a measurable
outcome which cannot be predicted ahead of time.
+ The probability model consists of three ingredients
Kenneth Harris
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1
A sample space S of possible outcomes of an experiment,
2
A collection E of events, which are subsets of S, and are said to
occur as the outcome of an experiment.
3
A probability measure P which assigns a nonnegative real
number to events (the “probability of the event occurring as the
outcome of an experiment").
Department of Mathematics
University of Michigan
January 19, 2009
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The probability model
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The Axioms of Probability
Events
Enter Tyche
+ Which subsets of the sample space S are events?
+ A probability measure on a sample space S is a function P defined
on the events E of the space S. We write
Discrete sample space. In a discrete sample space, we can safely
take any subset of the sample space S to be an event, which has
some probability of occurring.
P (E)
Nondiscrete sample space. Not all subsets of a nondiscrete sample
space can be events (which have some probability of occurring). This
issue will never be a problem for us in this class. It is actually hard to
construct such events in the continuous sample spaces in which we
will be most interested.
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to mean the probability of the event E occuring
Sample space. Whenever we talk about a probability measure P , we
are assuming it is a measure with respect to some given sample space
S.
+ A probability measure must also satisfy the three axioms which
follow.
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The Axioms of Probability
The Axioms of Probability
Axiom 1
Axiom 2
Axiom (1 – Nonnegative)
Axiom (2 – Unit measure)
For any event E, P (E) is a real number and
P (S) = 1.
0 ≤ P (E).
+ This axiom simply states that P (E) is a nonnegative real number for
any event E.
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+ The certain event S must always occur, and is defined to have
probability 1.
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The Axioms of Probability
Math 425 Introduction to Probability Lecture 5
Extended Addition Law
Axiom (3 – Addition rule (weak form))
There is nothing special about two events.
For any mutually exclusive events E and F (so, E ∩ F = ∅),
Proposition (Extended Addition Law)
If events E1 , E2 , . . . , En are mutually exclusive (so, Ei ∩ Ej = ∅
whenever i 6= j), then
P (E ∪ F ) = P (E) + P (F ).
P (E1 ∪ E2 ∪ . . . ∪ En ) = P (E1 ) + P (E2 ) + . . . + P (En ).
+ This version of Axiom 3 is most relevant when the sample space S
is finite.
Compare this axiom to the Sum Rule for counting:
In shorthand,
P
If two events E and F are mutually exclusive and E has n
outcomes and F has m outcomes, then E ∪ F has n + m
outcomes.
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The Axioms of Probability
Axiom 3 – Weak form
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Kenneth Harris (Math 425)
n
[
n
X
Ei =
P (Ei ).
i=1
i=1
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The Axioms of Probability
The Axioms of Probability
Proof
Axiom 3 – Strong form
I’ll prove the case of n = 3. Let E, F and G be mutually exclusive
events. Then E ∪ F and G are also mutually exclusive. (Why?)
Axiom (3 – Addition rule (strong form))
For any sequence of mutually exclusive events E1 , E2 , . . . (so,
Ei ∩ Ej = ∅ whenever i 6= j),
+ Apply the Addition Rule (Axiom 3):
P E ∪ F ∪ G = P (E ∪ F ) ∪ G
P(
= P (E ∪ F ) + P (G)
∞
[
Ek ) =
k =1
∞
X
P (Ek ).
k =1
= P (E) + P (F ) + P (G)
+ This rule is only relevant when the sample space S is infinite. The
Strong form of the Addition Rule really is stronger than its Weak form.
(See Ross, p. 30).
+ The n-event case is proved by induction, in the same way.
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Consequences of the Axioms
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Consequences of the Axioms
Probability of the Impossible Event
Proposition 4.1
Recall. ∅ is the impossible event.
Proposition (Ross 4.1)
For any event E,
Proposition
P (E c ) = 1 − P (E).
P (∅) = 0.
Proof.
Both S = E ∪ E c and E ∩ E c = ∅. So,
Proof.
Both S = S ∪ ∅ and S ∩ ∅ = ∅.
Apply the Addition Rule (Axiom 2):
1 = P (S)
= P (E ∪ E )
= P (E) + P (E c )
P (S) = P (S ∪ ∅) = P (S) + P (∅).
Axiom 3
So, P (E c ) = 1 − P (E).
So, P (∅) = 0.
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Axiom 2
c
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Math 425 Introduction to Probability Lecture 5
Consequences of the Axioms
Consequences of the Axioms
Proposition 4.2
Proof of Proposition 4.2
Proof.
Since E ⊆ F , we can express F as (see next slide)
Proposition (Ross, 4.2)
F = E ∪ (E c ∩ F )
Let E and F be any events. If E ⊂ F then P (E) ≤ P (F ).
where E and E c ∩ F are mutually exclusive.
By the Addition Rule
As a consequence, 0 ≤ P (E) ≤ 1 for any event E.
P (F ) = P (E) + P (E c ∩ F ) ≥ P (E),
where the last is because P (E c ∩ F ) ≥ 0 by Axiom 1.
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Consequences of the Axioms
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Consequences of the Axioms
Partition for Proposition 4.2
Proposition 4.3
Principle. Let E and F be any events in the same sample space.
Then E = E ∩ F and E c ∩ F are mutually exclusive, and
F = (E ∩ F ) ∪ (E c ∩ F ) = E ∪ (E c ∩ F ).
+ The following is the simplest version of the Inclusion-Exclusion
Identity. (See Ross, Proposition 4.4. This is for a later lecture.)
S
F
Proposition (Ross 4.3)
E
Let E and F be any events. Then
EÝF
P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F )
Ec Ý F
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Consequences of the Axioms
Consequences of the Axioms
Proof of Proposition 4.3
Partition for Proposition 4.3
+ By the Addition Rule (see next slide)
Principle. Let E and F be any events in the same sample space.
Then each of E ∩ F , E c ∩ F and E ∩ F c are mutually exclusive, and
P (E ∪ F ) = P (E ∪ F c ) + P (E c ∪ F ) + P (E ∩ F ).
E ∪ F = (E ∩ F ) ∪ (E c ∩ F ) ∪ (E ∩ F c ).
S
+ On the other hand, the Addition Rule gives (see next slide)
P (E) = P (E ∪ F c ) + P (E ∩ F )
P (F ) = P (E c ∪ F ) + P (E ∩ F )
E Ý Fc
EÝF
Ec Ý F
So,
E
P (E) + P (F ) − P (E ∩ F )
=
P (E ∪ F c ) + P (E c ∪ F ) + 2 · P (E ∩ F )
=
P (E ∪ F c ) + P (E c ∪ F ) + P (E ∩ F )
=
P (E ∪ F )
F
−P (E ∩ F )
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Consequences of the Axioms
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Consequences of the Axioms
Example 1
Example 1 – solution
Solution. Consider the events:
A: customers who carry AMEX,
V : customers who carry VISA.
Example
A retail store accepts VISA and AMEX. It has found that 24 percent of
its customers carry AMEX, 61 percent carry VISA and 11 percent carry
both.
What percentage of its customers carry VISA or AMEX?
Then, from the data
P (A) = 0.24 P (V ) = 0.61 P (A ∩ V ) = 0.11
Apply Proposition 4.4,
P (A ∪ V ) = P (A) + P (V ) − P (A ∩ V ) = 0.24 + 0.61 − 0.11 = 0.74.
+ So, 74 percent of the customers carry a credit card.
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Consequences of the Axioms
Consequences of the Axioms
Example 2
Example 2 – solution
Solution. Consider the events:
G: students who are Greek
I: students involved in intramural sports.
Example
Then, from the data (since Gc ∩ I c = (G ∪ I)c )
Sixty percent of students at a certain school are neither Greek nor
participate in an intramural sports. Twenty percent are Greek and 30
percent participate in intramural sports.
P (G ∪ I)c = 0.6
P (G) = 0.2
P (I) = 0.3
(a). By Proposition 1.1
(a) What is the percentage of students who are Greek or participate in
intramural sports?
(b) What is the percentage of student who do both?
P G ∪ I = 1 − P (G ∪ I)c = 1 − 0.6 = 0.4
So, (a) 40 percent of student are either Greek or participate in
intramural sports.
(b). Apply Proposition 4.3,
P (G ∩ I) = P (G) + P (I) − P (G ∪ I) = 0.2 + 0.3 − 0.4 = 0.1
So, (b) 10 percent of student participate in both activities.
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Finite sample space with equally likely outcomes
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Example 1
+ Let S is a finite sample space, and suppose it has N outcomes,
listed as
S = {a1 , a2 , . . . , aN }
Example
+ Then it follows from the Axioms that:
Solution. There are 36 possible outcomes on a pair of dice (any
number 1 to 6 could appear on each die). The throws with the same
face are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
For every i ≤ N,
1
P {ai } = ,
N
For any event E,
+ The desired probability is
|E|
,
N
where |E| is the number of outcomes in E.
P (E) =
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Assume all outcomes of a throw of a pair of dice are equiprobable.
What is the probability of rolling the same value on both die?
+ Suppose every outcome is equiprobable:
P {a1 } = P {a2 } = . . . = P {aN }
2
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Finite sample space with equally likely outcomes
Finite sample spaces, equiprobable outcomes
1
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6
36
= 16 .
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Finite sample space with equally likely outcomes
Finite sample space with equally likely outcomes
Example 2
Example 2 – solution
Solution. There are 25 possible outcomes.
+ Let Ei be the event that the ith toss is heads. Then |Ei | = 24 (fix ith
toss at H). So,
24
1
P (Ei ) = 5 = .
2
2
Example
A coin is tossed five times in succession. Assume each sequence of
heads/tails is equiprobable.
What is the probability of getting heads on the first or third toss?
+ For i 6= j, |Ei ∩ Ej | = 23 (fix ith and j toss at H). So,
P (Ei ∩ Ej ) =
+ The sample space S for this problem are sequences
(a1 , a2 , a3 , a4 , a5 )
1
23
= .
5
2
4
+ Apply Proposition 4.4:
where each ai = H or T
P (E1 ∪ E3 ) = P (E1 ) + P (E3 ) − P (E1 ∩ E3 ) =
For example: (H, T , H, T , H).
1 1 1
3
+ − = .
2 2 4
4
+ The desired probability is 34 .
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Finite sample space with equally likely outcomes
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Finite sample space with equally likely outcomes
Example 3
Example 3 – solution
Example
Solution. The sample space of the Pick-Six Lottery are all subsets of
{1, 2, 3, . . . , 49} with six elements:
49
possible outcomes .
6
Some states have the Pick-Six Lottery: A person purchases a ticket,
and can choose 6 distinct numbers in the set {1, 2, 3, . . . , 49}.
Later a Lottery Machine picks 6 distinct numbers at random in the
set {1, 2, 3, . . . , 49}.
A winning ticket is one which matches the six numbers chosen by
the Machine (in any order of selection).
+ By assumption, all possible selections of numbers are equiprobable,
so the likelihood that a ticket is a winner is
1
=
49
6
+ What are the odds of winning with one ticket?
1
.
13, 983, 816
+ You have better odds getting 23 consecutive tosses of heads on a
fair coin (where heads/tails are equally likely on each toss).
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