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STAT151: Introduction to Statistical Theory
Term I, 2013-2014
Yang Zhenlin
[email protected]
http://www.mysmu.edu/faculty/zlyang/
Course Contents
Chapter 1
This course serves as an introduction to basic statistical theory for
students who intend to pursue a quantitative major at SMU such as
Applied Statistics, Actuarial Science, Economics, Finance, Marketing,
Management, etc. Students are expected to have some basic knowledge
in probability and statistics and to be mathematically oriented or at the
very least, be interested in mathematics. Major topics include:
Probability and Conditional Probability
Distributions and Conditional Distributions
Point Estimations
Confidence Intervals
Testing Statistical Hypotheses
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Course Outline
© Zhenlin Yang, SMU
Chapter 1: Probability
Chapter 1
In any scientific study of a physical phenomenon, it is
desirable to have a mathematical model that makes it
possible to describe or predict the observed value of some
characteristic of interest.
The physical situations may be deterministic, such as the
speed of a falling object in a vacuum, travel distance of an
airplane in certain time period when the speed is set to a
constant, etc., or may be nondeterministic, such as the
outcome of tossing a coin, the lifetime of a light bulb, etc.
This course concerns the latter situation: to provide a
probability model so as to facilitate the “statistical inference”
for the nondeterministic situation.
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Brief History
Chapter 1
Concepts of probability have been around for thousands of years, but
probability theory did not arise as a branch of mathematics until the
mid-seventeenth century.
In the mid-seventeenth century, a simple question directed to Blaise
Pascal by a nobleman sparked the birth of probability theory.
Chevalier de Méré gambled frequently to increase his wealth. He bet
on a roll of a die that at least one 6 would appear during a total of 4
rolls, and was more successful than not with this game of chance.
Tired of his approach, he decided to change the game. He bet on a
roll of two dice that he would get at least one double 6, on twentyfour rolls of two dice. Soon he realized that his old approach to the
game resulted in more money. He asked his friend Blaise Pascal
why his new approach was not as profitable.
Pascal worked through the problem and found that the probability of
winning using the new approach was only 49.14 percent compared to
51.77 percent using the old approach.
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Brief History
Chapter 1
This problem proposed by Chevalier de Méré is said to be the start of
the famous correspondence between Blaise Pascal and Pierre de
Fermat. They continued to exchange their thoughts on mathematical
principles and problems through a series of letters.
Historians think that the first letters written were associated with the
above problem and other problems dealing with probability theory.
Therefore, Pascal and Fermat are
the mathematicians credited with
the founding of probability theory
(David, 1962).
The topic of probability is seen in many facets of the modern world. The
theory of probability is not just taught in mathematics courses, but can be
seen in practical fields, such as insurance, industrial quality control, study of
genetics, quantum mechanics, and the kinetic theory of gases (Simmons,
1992).
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Learning Objectives
Chapter 1
Sample Space and Events
Basic Set Algebra
Definition of Probability
Conditional Probability
Rules for Calculating Probabilities
Probability Trees
Counting Techniques
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Chapter 1
Sample Space and Events
Some fundamental concepts are important to the learning of
probability theory:
An experiment refers to the process of obtaining an
observed result of some phenomenon.
A trial refers to a run of an experiment.
An outcome refers to the observed result.
The set of all possible outcomes of an experiment is called
the sample space, denoted by S.
Note: One and only one of the possible outcomes will occur on
any given trial of the experiment.
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Chapter 1
Sample Space and Events
Example 1.1 An experiment consists of tossing two coins, and the
observed face of each coin is of interest. The set of possible
outcomes is represented by the sample space:
S = {HH, HT, TH, TT}, where H=Head, and T=Tail.
Example 1.2. Suppose in Example 1.1, we are instead interested
in the total number of heads obtained from the two coins. An
appropriate sample space could then be defined as
S = {0, 1, 2}.
Thus, a different sample space may be appropriate for the same
experiment, depending on the characteristic of interest.
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Sample Space and Events
Chapter 1
More fundamental concepts:
A sample space S is said to be finite if it consists of a finite
number of elements, say S = {e1, e2, . . . , eN}, and countably
infinite if its outcomes can be put into a one-to-one
correspondence with the positive integers 1, 2, 3, . . .
If a sample space S is either finite or countably infinite, then it
is called a discrete sample space; If the outcomes can assume
any value in some interval of real numbers then the sample
space is called a continuous sample space.
Note: the sample spaces in Examples 1.1 and 1.2 are discrete.
Example 1.3. If a coin is tossed repeatedly until a head occurs,
then a natural sample space is
S = {H, TH, TTH, TTTH, . . . }.
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Sample Space and Events
Chapter 1
If in Example 1.3 one is interested in the number of tosses
required to obtain a head, then a possible sample space would
be the set of all positive integers, i.e., S = {1, 2, 3, . . . }.
Clearly these are countable and infinite sample spaces.
Example 1.4. A light bulb is placed in service and the time of
operation (in hours) until it burns out is measured, then the
sample space consists of all nonnegative real numbers, i.e.,
S = {t | 0 ≤ t < ∞},
which is a continuous sample space.
Event. Any subset A of the sample space S is defined as an
event. If the outcome of an experiment is contained in A, then
we say event A has occurred in this experiment.
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Set Algebra
Chapter 1
The study of probability models requires a familiarity of
basic notations of set theory.
A set is a collection of distinct objects. Sets usually are
designated by capital letters, A, B, C, . . . , or subscripted
letters A1, A2, A3, . . .
Individual objects in a set A are called elements.
In the context of probability, the sets are called events and the
elements are called outcomes.
Universal set S is the set of all elements under consideration.
In probability application it is called the sample space.
Empty set or null set, denoted by , is the set that contains no
element. In probability, it is empty event.
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Chapter 1
Set Algebra
If all the elements in a set A also are contained in another set
B, we say that A is a subset of B, denoted by A  B.
It is always true that   A  S.
Some of these sets are depicted using figures called the
Venn diagrams:
S
S
A
A
Event A: the shaded area
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B
Event B is a subset of Event A
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Chapter 1
Set Algebra
The union of two sets A and B, denoted by AB, is a new set
consisting of all elements that are either in A or in B or in
both A and B, i.e.,
AB = {a | a  A or a  B}
The intersection of A and B, denote by AB, is a new set
containing all elements that are both in A and in B, i.e.,
AB = {a | a  A and a  B }
S
S
AB
AB
AB: the shaded area
AB: the shaded area
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Chapter 1
Set Algebra
The complement of A, denoted by Ac, consists of all outcomes
in S that are not in A;
The difference of A and B, denoted by A – B, is defined as a
new set that consists of all elements in A but not in B.
Clearly, A – B = ABc.
S
A
A
Ac: the shaded area
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B
A-B: the darker shaded area
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Chapter 1
Set Algebra
The notions of union and intersection can be extended to more
than two sets. For example,
ABC = {a | a  A or a  B or a  C }
ABC = {a | a  A and a  B and a  C}
A
C
ABC: all shaded areas
B
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ABC: the darkest area
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Chapter 1
Set Algebra
Tow sets A and B are said to be disjoint if they have no
elements in common. In probability context they are called
mutually exclusive events.
S
S
A
B
A
A and B are not mutually exclusive
B
A and B are mutually exclusive
Example 1.5. An experiment consists of tossing two dice. The
sample space thus consists of 36 points: {(1,1), (1,2), (1,3),
(1,4), (1,5), (1,6), . . . , (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}, or
in short,
S = {(i, j) | i, j = 1, 2, 3, 4, 5, 6}
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Chapter 1
Set Algebra
Let A be the event that the sum of the dice equals to 7, and B be
the event that the first die shows a 5. Then
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Now,
AB = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1),
(5, 1), (5, 3), (5, 4), (5, 5), (5, 6)}
AB = {(5, 2)}
A–B = {(1, 6), (2, 5), (3, 4), (4, 3), (6, 1)}
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Chapter 1
Set Algebra
Important laws of sets or events:
A(BC) = (AB)C and A(BC) = (AB)C
(Associative Law)
AB = BA and AB = BA (Commutative Law)
A(BC) = (AB)(AC) and
A(BC) = (AB)(AC) (Distribution Law).
A = A,
AS = A,
AAc = S, and
AAc = .
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Set Algebra
Chapter 1
Other obvious but useful equations include:
• (Ac) c = A,
• c = S, Sc = ,
• AA = A, AA = A, AS = S,
• A = ,
• A(AB) = A,
• A(AB) = A.
De Morgan's law: (AB)c = AcBc and (AB)c = AcBc
(AB)c = Green Area
AcBc = Green Area
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Set Algebra
Chapter 1
Mutually Exclusive Events and Partition of Sample Space
Events A1, A2, A3, . . . , are said to be mutually exclusive if
they are pairwise mutually exclusive. That is, Ai Aj = 
whenever i ≠ j.
If A1, A2, . . . , Ak are mutually exclusive and the union of them
makes up the sample space S, then, A1, A2, . . . , Ak are said to
form a partition of the sample space S, i.e.,
S = A1A2 . . . Ak
Clearly for any event B, the events BA1, BA2, . . . , BAk
form a partition of the event B, in the sense that
B = (BA1)  (BA2)  . . .  (BAk).
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Chapter 1
Set Algebra
See Venn diagrams below for the case of k = 6.
This demonstration is useful in introducing a very important
law: the Law of Total Probability.
A2
A3
A1
A2
A4
A6
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A1
A5
A6
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A3
A4
A5
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Definition of Probability
Chapter 1
For a given experiment with a sample space S, the primary
objective of probability modeling is to assign to each event A an
real number P(A), called probability of A, that will provide a
measure of the likelihood that A will occur when the
experiment is performed.
The probability P(A) of a event A in a sample space S is defined
as a set function with domain being a collection of events, and
range a set of real numbers, such that
Axiom 1:
0 ≤ P(A) ≤ 1 for every A  S
Axiom 2:
P(S) = 1

Axiom 3:
P(i1 Ai )  i 1 P( Ai ) , if A1, A2, . . . , are
pairwise mutually exclusive.
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Chapter 1
Interpretation of Probability
Classical Interpretation of Probability
The oldest way of measuring uncertainties is the classical
probability concept, developed originally from games of
chance and applies when all possible outcomes are equally
likely to occur:
If there are N equally likely possibilities, of which one must
occur and n(A) are contained in event A, then the probability
of A is given by P(A) = n(A)/N.
Using Venn
diagram, the
geometric
interpretation:
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S
Area of A
P( A) 
Area of S
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A
© Zhenlin Yang, SMU
Chapter 1
Interpretation of Probability
Example 1.6. What is the probability of drawing an ace from a well-shuffled deck
of playing cards?
Solution: there are n(A) = 4 aces, and N = 52 cards, so we get
P(A) =
n(A)
4
1
=
 .
N
52 13
Frequency Interpretation of Probability
A major shortcoming of the classical probability concept is its
limited applicability, for which there many situations in which
the various possibilities cannot be regarded as equally likely.
Among the various probability concepts, most widely held is
the frequency interpretation:
The probability of an event is the proportion of the time it
occurs in the long run when the experiment is repeated.
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Interpretation of Probability
Chapter 1
For example, we say that a coin is fair if repeated tossing results
in about 50% heads.
Probability as a Measure of Belief
An alternative point of view is to interpret the probabilities as
personal or subjective evaluations. Such probabilities
express the strength of one's belief with regards to the
uncertainties involved, and they apply especially when there
is little or no direct evidence, such as risk taking and betting
situations.
For example, in a 5-horse race, you feel that the first horse has
30% chance of winning, 2nd horse 25%, 3rd 20%, 4th 15%,
and 5th 10%.
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Rules for Calculating Probabilities
Chapter 1
Complement Rule: If A is an event and Ac is its complement, then
P(A) = 1 – P(Ac)
Note: This result is particularly useful when an event is relatively
complicated, but its complement is easier to handle.
Example 1.7. An experiment consists of tossing a coin four
times. What is the probability of obtaining at least one head?
Solution: Here the event of interest is A = 'at least one head.'
This event is complicated, but Ac = 'no heads' = {TTTT}, which
contains only one outcome. Thus
P(A) = 1 – P(Ac) = 1 – P(TTTT)
15
1
= 1 – 24 =
16
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Chapter 1
Rules for Calculating Probabilities
Addition Rule: For any two events A and B, the following is true:
P(AB) = P(A) + P(B) – P(AB)
This can be seen using the Venn
diagram on left where E1 = Green,
E2 = Purple, and E3 = Dark Green.
S
E1
A
Note E3 = AB, we have,
E3
E2
B
P(AB) = P(E1E2E3) = P(E1) + P(E2) + P(E3)
= [P(E1) + P(E3)] + [P(E2) + P(E3)] – P(E3)
= P(A) + P(B) – P(AB)
However, P(A)+P(B) = [P(E1)+P(E3)]+ [P(E2)+P(E3)], where
P(E3) = P(AB), is added twice,
P(AB)  P(A) + P(B)!
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Rules for Calculating Probabilities
Chapter 1
Random Selection: if an object is chosen from a finite
collection of distinct objects in such a manner that each object
has the same probability of being chosen, then we say that the
object was chosen at random.
Example 1.8. Suppose one card is drawn at random from a wellshuffled deck of playing cards. Find the probability that the card
is either an ace or a heart.
Solution: Since the card is selected at random, each card thus
has the same probability, 1/52, of being chosen. Let A = "select
an ace", and B = "select a heart". Then
P(A or B) = P(AB) = P(A) + P(B) – P(AB)
= 4/52 + 13/52 – 1/52 = 16/52 = 4/13.
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Conditional Probability
Chapter 1
The most important concept in probability may be the
conditional probability. This is because
Partial information concerning the result of experiment often
is available. Given the known information, one certainly
knows better whether an event can occur or not; and
Calculation of probability of certain event sometimes can only
be done through conditioning.
The conditional probability of an event A, given that another
event, B, has occurred, is defined by
P( A  B)
P( A | B) 
P( B)
provided that P(B) ≠ 0.
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Chapter 1
Conditional Probability
Geometrical Interpretation of P(A|B):
P( A  B)
P( A | B) 
P( B)
Area of ( A  B)

Area of B
S
A
AB
B
 the magnitude of P(A|B) depends on the relative sizes of the
event AB and the event B.
Area of A
P( A) 
Area of S
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 the magnitude of P(A) depends
on the relative sizes of the event
A and the sample space S.
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Chapter 1
Conditional Probability
S
A
S
A
B
B
AB
0  P(A|B)  1
P(A|B) = 0
S
A
AB
S
B
A
AB = B
P(A|B) = 1
0  P(A|B)  1
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Conditional Probability
Chapter 1
“Conditioning on B” can be understood as reducing the sample
space from S to B. It is easy to see that conditional probability
satisfies the conditions of a probability set function, hence it
enjoys all the properties of probability, e.g.,
(1) P(Ac|B) = 1 – P(A|B)
(2) P(A1A2|B) = P(A1|B) + P(A2|B) – P(A1A2|B)
Example 1.9. A box contains 100 microchips, 60 were produced
by factory 1 with 15 defectives, and the rest by factory 2 with 5
defectives. One microchip is selected at random from the box
and tested. What is the probability that the microchip was
produced by factory 1 if the test indicates that it is a good one?
What if the test indicates that it is a defective one?
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Chapter 1
Conditional Probability
Solution. Let A = {the microchip was produced by factory 1}
B = {obtaining a good microchip}
The desired probabilities are P(A B) and P(A Bc ) .
Method 1: direct calculation:
Of 80 good ones, 45 from factory 1, hence P(A B) = 45/80 = 9/16
Of 20 defective ones, 15 from factory 1, hence P(A Bc ) = 15/20 = 3/4
Method 2: from definition.
P(A B) =
P(A  B) 45 100
9
=
=
P(B)
80 100 16
c
P(A

B
)
15 100 3
c
=
=
P(A B ) =
P(Bc )
20 100 4
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Conditional Probability
Chapter 1
Example 1.10. A pair of four-sided dice is rolled and the sum is determined. What
is the probability that a sum of 3 is rolled before a sum of 5 is rolled?
Solution: Define A = {sum=3} and B = {sum=3 or 5}.
probability is P(A B) (why?),
Then the desired
which can be calculated directly as:
P(A  B)
P(A B) =
P(B)
=
P(A) 2 16 1
=
= .
P(B) 6 16 3
A direct application of the definition of the conditional probability gives the
following important result.
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Chapter 1
More Probability Rules
Law of Multiplication: For any two events A and B,
P(AB) = P(A|B)P(B) = P(B|A) P(A)
Example 1.11. At a fair a vendor has 25 helium balloons on
strings: 10 balloons are yellow, 8 are red, and 7 are green.
Balloons are sold in random order. What is the probability that
the first two balloons sold are both yellow?
Solution: Let A = {the first balloon sold is yellow},
B = {the second balloon sold is yellow}.
The quantity of interest is
P(AB) = P(A)P(B|A)
= (10/25)(9/24) = 3/20.
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More Probability Rules
Chapter 1
Extended Law of Multiplication: For any three events A, B
and C
P(ABC) = P(AB)P(C|AB) = P(A)P(B|A) P(C|AB),
For any four events A, B, C and D,
P(ABCD) = P(A) P(B|A)P(C|AB) P(D|AB C),
etc.
Sampling with replacement occurs when an object is
selected and then replaced before the next object is selected;
Sampling without replacement occurs when an object is
selected and not replaced before the next object is selected.
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More Probability Rules
Chapter 1
Example 1.12. Four cards are drawn at random without replacement from a deck
of 52 cards. Find the probability of getting "an ace on the first draw", "a king on
second draw", "an ace again on third draw", and "a queen on the last draw".
Solution: Define
A = {getting an ace on the first draw}
B = {getting an king on the 2nd draw}
C = {getting an ace on the 3rd draw}
D = {getting an queen on the 4th draw}
The desired probability is P(A  B  C  D) that can be easily calculated by the
Law of Multiplication:
P(A  B  C  D) = P(A) P(B A) P(C A  B) P(D A  B  C)
 4  4  3  4 
=      .
 52  51 50  49 
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Chapter 1
Independent Events
Two events A and B are called independent events if
P(AB) = P(A) P(B),
Otherwise, A and B are called dependent events.
It follows from the definition of independence that A and B
are independent if and only if either of the following holds:
P(A|B) = P(A) or P(B|A) = P(B);
It follows from the definition of mutually exclusiveness that,
A and B are mutually exclusive events if and only if either of
the following holds:
P(A|B) = 0 or P(B|A) = 0.
Hence, independence and mutually exclusiveness are very
much different.
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Independent Events
Chapter 1
Example 1.13. A box contains eight tickets, each labeled with a binary number.
Two are labeled 111, two 100, two 010, and two 001.
111
100
010
001
111
100
010
001
An experiment consists of drawing one ticket at random from the box. Let A be
the event that the "the first digit is 1", B be the event that "the second digit is 1",
and C be the event that "the third digit is 1". Are events A and B independent?
Are A and C independent? Are B and C independent?
Solution: From the illustration above we have
P(A) = P(B) = P(C) = 4/8 = 1/2
P(A  B) = P(A  C) = P(B  C) = 2/8 =1/4
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Chapter 1
Independent Events

Hence A and B are independent, so are the events A and C, and B
and C. Such events are called pairwise independent events.
Three events A, B and C are said to be independent if they are
pairwise independent and P(ABC) = P(A)P(B)P(C).
Example 1.13. (Cont'd). Clearly, the three events are not
totally independent since
P(A B C) = 2/8 ≠ P(A)P(B)P(C) = 1/8.
Pairwise independence does not imply independence;
but independence implies pairwise independence.
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Chapter 1
More Probability Rules
Sometimes it is not possible to calculate P(A) directly, but may be
possible to calculate it part by part and then add up.
Law of Total Probability: If A1, A2, … , Ak is a collection of
mutually exclusive events such that the union of them makes up
the sample space, then for any event B, we have,
k
k
i 1
i 1
P( B)   P( Ai  B)   P( B | Ai ) P( Ai )
A2
The Venn diagram on the
right illustrates the case of
6 events, which form a
partition of sample space.
STAT151, Term II 14-15
A1
A6
41
A3
B
A4
A5
© Zhenlin Yang, SMU
Chapter 1
More Probability Rules
Example 1.14. Suppose that 80% of the seniors, 70% of the juniors, 50% of the
sophomores, and 30% of the freshmen of a college use the library of their campus
frequently. If 30% of all students are freshmen, 25% are sophomores, 25% are
juniors, and
frequently?
20%
are seniors.
What percent of all students use the library
Solution: To solve this type of problems, follow the steps:
Step 1: Define events involved using proper letters:
A = {a randomly selected student uses library frequently}
F, O, J, and E be the events that he or she is a freshmen, sophomore,
junior, or senior, respectively.
Step 2: Write out the given information:
P(F) = .30,
P(O) = .25, P(J) = .25,
P(A F) = .3,
P(A O) = .5, P(A J ) = .7,
STAT151, Term II 14-15
42
P(E) = 0.20
P(A E) = .8.
© Zhenlin Yang, SMU
Chapter 1
More Probability Rules
Step 3: Express the question(s):
The desired probability is P(A).
Step 4: Identify the appropriate method to solve the problem:
The Law of Total Probability.
P(A) = P(A F) P(F)+ P(A O) P(O)+ P(A J ) P(J)+ P(A E) P(E)
= (.3)(.3) + (.5)(.25) + (.7)(.25) + (.8)(.20) = .55
The next result will help answer the question: Given B has
happened, was it due to A1 or A1 or … or Ak ?
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
More Probability Rules
Chapter 1
Bayes' Theorem: If the events A1, A2, … , Ak form a partition
of the sample space S, then for any event B in S, we have,
P( Ai | B) 
P( B | Ai ) P( Ai )
k
 P( B | A ) P( A )
i
i 1
i
In practice, the probabilities P(Ai) are called prior probabilities,
and the conditional probabilities are called posterior
probabilities.
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
More Probability Rules
Chapter 1
Example 1.15. Bowl 1 contains two red and four white chips; bowl 2 contains one
red and two white chips; and bowl 3 contains five red and four white chips.
Suppose the probabilities for selecting the bowls are not the same but are given by
P(B1) = 1/3, P(B2) = 1/6, P(B3) = 1/2, where B1, B2, and B3 are the events that
bowls 1, 2, 3 are chosen, respectively. The experiment consists of selecting a bowl
with these probabilities and then draw a chip at random from that bowl. What is
the probability of drawing a red chip? Given that a red chip is obtained, what is the
probability that it came from bowl 1, 2, or 3, respectively?
Solution: Let R represent the event of getting a red chip.
P(R) = P(R|B1) P(B1) + P(R|B2) P(B2) + P(R|B3) P(B3)
4
 1  2   1  1   1  5 
=            =
9
 3  6   6  3   2  9 
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
More Probability Rules
P(B1 R)
=
2
(1 / 3)( 2 / 6)
P( B1  R)
=
=
8
4/9
P( R)
P(B2 R) =
1
(1 / 6)(1 / 3)
P( B2  R)
=
=
8
4/9
P( R)
P(B3 R) =
5
P( B3  R)
= (1/ 2)(5 / 9) = .
8
P( R)
4/9
Chapter 1
Given that a red chip is
obtained, it is most likely
that it came from bowl 3.
and
In applying the Law of Total Probability and Bayes’ Rule, one
needs to identify two sets of given information: one is a set of
unconditional probabilities, and the other is a set of conditional
probabilities.
The key words such as ‘given’, ‘of’ and ‘among’ indicate that
the subsequent number is a conditional probability.
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
When an experiment consists of several operations and each
operation has a certain number of possible outcomes, then the
total number of possible outcomes for the experiment can be
found from the following principle.
Counting Principle: Let E1, E2, ..., Ek be sets with n1, n2, ...,
nk elements, respectively. Then there are n1 n2 …  nk ways in
which we can first choose an element of E1, then an element of
E2, ..., and finally an element of Ek.
The Counting Principle can be used to solve many probability
problems when sample space is finite and the outcomes are all
equally likely.
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
Example 1.16. Suppose that there n routes from a town, A, to
another town, B, m routes from town B to a third town, C,
and l routes from town C to a fourth town, D. If we decided to
go from A to D via B and C. How many different routes are
there from A to D?
Solution: In this situation, there are k = 3 operations: A to B, B to C,
and C to D, with number of elements (different routes) n, m, and l,
respectively. Hence the total number of different routes is n  m  l.
The above can be argued in a direct way. For each route that we
choose from A to B, we have m choices from B to C. Therefore,
altogether we have nm choices to go from A to C via B.
Furthermore, for each choice from A to C, there are l choices from
C to D. Hence we totally have nml choices from A to D.
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
Example 1.17. (Birthday Problem) What is the probability that at least two
students of a class of size n have the same birthday? Assume that the birth rates are
constant and there are 365 days a year. Compute the numerical values of such
probabilities for n = 23, 30, 50, and 60.
Solution: There are 365 possibilities for the birthdays of each of the n students.
Therefore, there are 365n possible combinations of the birthdays of n students. Let
A = {no two students have the same birthday}, then
365  364    [365  (n  1)]
P(A) =
365 n
The desired probability is thus, P(Ac ) = 1– P(A).
Now, n = 23, P(Ac ) = 0.507; n = 30, P(Ac ) = 0.706; n = 50, P(Ac ) = 0.970; n = 60,
P(Ac ) = 0.995.
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Chapter 1
Counting Techniques
Permutation: An ordered arrangement of r objects from a
set A containing n objects (0 ≤ r ≤ n) is called an r–element
permutation of A or a permutation of the elements of A
taken r at a time.
For example, if A = {a, b, c, d}, then ab and ba are two-element
permutations of A; bcd, bdc, cbd, cdb, dbc and dcb are the threeelement permutation of A.
Permutation Rule: Let nPr denote the total number of
permutations of a set A containing n elements taken r at a
time. Then
n!
nPr = n(n – 1)(n – 2) ... (n –r + 1) =
(n  r )!
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
Example 1.18. In how many different ways can a president, a
treasure, and a secretary be selected from a group of 20 people to
form a committee?
Solution: nPr = 20P3 = 201918 = 6840.
–––There are 20 ways to choose a president, 19 ways of choosing a treasure
after a president has been chosen, and 18 choices left for a secretary.
Example 1.19. If five boys and five girls sit in a row in a random
order, what is the probability that no two children of the same sex
sit together?
Solution: First we know that there are 10! possible seating patterns in total for
the 10 children. In order that no two children of the same sex sit together, boys
must occupy positions 1, 3, 5, 7 and 9 and girls 2, 4, 6, 8 and 10.
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
There are 5! different ways for the five boys to occupy
positions 1, 3, 5, 7, 9, and 5! different ways for the five girls to
occupy positions 2, 4, 6, 8 and 10.
Hence the total number of different ways for this to happen is
5!5!
Boys and girls can switch positions, i.e., boys occupy positions
2, 4, 6, 8 and 10 and girls 1, 3, 5, 7 and 9,
Hence the total number of seating patterns such that no two
children of the same sex sit together is 25!5!
Therefore, the desired probability is
2  5!5!
 0.08
10!
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
Combination: An unordered arrangement of r objects from
a set A containing n objects (0 ≤ r ≤ n) is called an r–element
combination of A or a combination of the elements of A
taken r at a time.
Therefore, two combinations are different only if they differ in composition;
but different permutations can have the same composition, e.g., if A = {a, b, c,
d}, then ab and ba give the same two-element combination, but different twoelement permutations of A.
Combination Rule: Let nCr or  nr  denote the total number of
combinations of a set A containing n elements taken r at a
time. Then
n!
n Pr
(read as: n choose r combination)

nCr =
r! r!(n  r )!
STAT151, Term II 14-15
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© Zhenlin Yang, SMU
Counting Techniques
Chapter 1
Example 1.20. In how many different ways can a committee of
three people be formed from a total of 20?
Solution: Since the order is immaterial, the answer is = 1140, i.e. there are
1140 different ways (smaller than the number in Example 1.18).
Example 1.21. A poker hand is defined as drawing 5 cards at
random without replacement from a deck of 52 cards. What is the
probability of obtaining a full house (one pair and one triple of
cards with equal face value)?
Solution: There are 13 denominations. Choose 2 denominations, and then
choose two cards from the first denomination and three from the second. Thus,
P(Full House) =
STAT151, Term II 14-15
13
P2 4 C2 4 C3 13 12  4  3  4 2

 0.00144
52! (47!5!)
52 C5
54
© Zhenlin Yang, SMU
Chapter 1
Counting Techniques
Probability Trees
Example 1.22. Find the probability of selecting two female students
(without replacement), if there are 3 female students in a class of 10.
Solution:
P(F|F) =2/9
P(FF) = P(F)P(F|F)
= (3/10)(2/9)
P(F) =3/10
Fisrt Selection
P(M) =7/10
P(M|F) =7/9
Second Selection
P(F|M) =3/9
P(M|M) =6/9
STAT151, Term II 14-15
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© Zhenlin Yang, SMU