Chapter 2. Foundations of Probability
Section 2.3. Definition of Probability
Jiaping Wang
Department of Mathematical Science
01/16/2013, Wednesday
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Outline
Introduction
Definition of Probability
Inclusive-Exclusive Principle
Homework #1
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Part 1. Examples
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Introduction
Intuitive idea of probability is related to relative frequency of occurrence.
For example, in flipping a coin, we will think the probability to obtain head is ½ and
same for obtaining tail.
Another example like rolling a die, the probability to obtain a number among
{1,2,3,4,5,6} would be 1/6 for each number, so that the probability to obtain
the even number is ½ = 1/6+1/6+1/6 and same for obtaining the odd number.
Kolmogorov proposed the three axioms to define the
probability in Foundations of the Calculus of Probabilities.
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Part 2. Definition of Probability
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Definition 2.3
Suppose that a random experiment has associated with a
sample space S. A probability is a numerically valued
function that assigned a number P(A) to every event A so
that the following axioms hold:
(1) P(A) ≥ 0
(2) P(S) = 1
(3) If A1, A2, … is a sequence of mutually exclusive events
(that is Ai∩Aj=ø for any i≠j), then
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Some Basic Properties
1. P( ø ) = 0, P(S) = 1.
2. 0≤ P(A) ≤1for any event A.
3. P(AUB) = P(A) + P(B) if A and B are mutually
exclusively.
4. P(AUB) = P(A) + P(B) – P(A∩B) for
general events A and B.
5. If A is a subset of B, then P(A) ≤ P(B).
6. P(A) = 1 – P(A).
7. P(A∩B) = P(A) – P(A∩B).
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Part 3. Inclusive-Exclusive
Principle
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Union of More Events
We know P(AUB) = P(A) + P(B) – P(A∩B), how about P(AUBUC)?
Consider a new event D=BUC, then
P(AUBUC) = P(A)+P(D)-P(A∩D)
=P(A)+P(BUC)-P((A∩B)U(A∩C))
=P(A)+P(B)+P(C) – P(B∩C) – [P(A∩B) +P(A∩C)
- P(A∩B∩A∩C)]
=P(A)+P(B)+P(C) – P(B∩C) – P(A∩B)- P(A∩C)
+ P(A∩B∩C)
How about P(AUBUCUD) and P(A1UA2UA3UA4U….. UAn)
for any integer n>0?
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Inclusive-Exclusive Principle
Theorem 2.1. For events A1, A2, …, An from the sample space S,
We can use induction to prove this.
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Determine the Probability Values
The definition of probability only tells us the axioms that the probability
function must obey; it doesn’t tell us what values to assign to specific event.
The value of the probability is usually based on empirical evidence or on
careful thought about the experiment.
For example, if a die is balanced, then we may think P(Ai)=1/6 for Ai={ i },
i = 1, 2, 3, 4, 5, 6
However, if a die is not balanced, to determine the probability, we need run
lots of experiments to find the frequencies for each outcome.
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Example 2.3
A farmer has decided to plant a new variety of corn on some of his land, and
he has narrowed his choice to one of three varieties, which are numbered 1,
2, and 3. All three varieties have produced good yields in variety trials. Which
corn variety produces the greatest yield depends on the weather. The optimal
conditions for each are equally to choose, the farmer writes the name of each
variety on a peace of paper, mixes the pieces, and blindly selects one. The
variety that is selected is purchased and planted. Let Ei denote the event that
variety i is selected (i = 1, 2, 3), let A denote the event that variety 2 or 3 is
selected, and let B denote the event that variety 3 is not selected. Find the
probabilities of Ei, A and B.
Solution: Sample space S={1,2,3}, so E1={1}, E2={2} and E3={3}. Intuitively, it is
reasonable to assign a probability 1/3 to each Ei, that is
P(E1)=P(E2)=P(E3)=1/3
And A=E2UE3 , thus, P(A)=P(E2)+P(E3)=2/3, and P(B)=1-P(E3)=1-1/3=2/3.
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Example 2.4
The National Crime Victimization Survey (NCVS) is an ongoing survey of a
national representative sample of residential households. Twice each year,
data are obtained from about 49,000 households, consisting of approximately
100,000 persons, on the frequency, characteristics, and consequences of
victimization in the United States. Results for 1995 and 2002 are given in the
following table.
19 95
20 02
Rates per 1000 persons
Ages 18-24
*Based on 10 or fewer
sample cases
College NonStudents Stude
nts
College NonStudent Stude
s
nts
Violent Crime
87.7
101.6
40.6
56.1
Rape/Sexual Assault
4.3
4.4
3.3
4.1
Robbery
8.4
12.1
2.9*
6.8
Aggravated Assault
14.5
22.2
9.1
13.2
Simple Assault
60.5
62.8
25.3
32.0
Serious Violent Crime 27.3
38.8
15.3
24.1
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Q&A
Q1: For a randomly selected college student in 1995, what is the probability that
the person was a victim of simple assault that year?
A1: 60.5/1000=0.0605.
Q2: For a randomly selected college student in 2002, what is the probability that
the person was a victim of simple assault that year?
A2: 25.3/1000=0.0253.
Q3: For a randomly selected non-student in 2002 who was in the 18 to 24 age group,
what is the probability that the person was not a victim of a violent crime that year?
A3: 1-56.1/1000=1-0.0561 = 0.9439.
Q4: Note that in table, the category “Violent Crime” has five subcategories, why do the
rates in the five subcategories not total to the Violent Crime rate?
A4: They are not mutually exclusively.
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More Example
In a class of 60 students, 13 could not roll their tongue, 17 had attached
earlobes and 10 could roll their tongues and had earlobes. A student is
randomly selected from this class. Find the probability that the selected
student:
1. Can roll his or her tongue;
2. Could not roll his or her tongue and had attached earlobes;
3. Could either roll his or her tongue or had attached earlobes but not both.
Sample space S={60 students}, A={students could roll tongue},
B={students had attached earlobes}. P(A)=13/60 , P(B)=17/60, P(A∩B)=10/60
Answer 1: P(A)= 1-13/60 =47/60= 0.783
Answer 2: P(A ∩B)= P(B)-P(A ∩B)=17/60-10/60=7/60=0.117
Answer3: P[(A ∩B)U(A ∩B)]=P(A ∩B)+P(A ∩B)=7/60+P(A)-P(A ∩B)=7/60+47/60-10/60
=44/60=11/15=0.733
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Homework #1
Page 20: 2.8, 2.10
Page 21: 2.14, 2.15
Page 29: 2.18, 2.20
Page 31: 2.27, 2.28
Due next Wed., 01/23/2013
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