Transcript A, B

CHAPTER 6: PROBABILITY PART I
Outline
•
•
•
•
•
•
Introduction
Probability rules and trees
Probability distributions
Expected value and variance
Binomial distribution
Poisson distribution
1
BENDRIX COMPANY’S SITUATION
• The Bendrix Company supplies contractors with materials
for the construction of houses.
• Bendrix currently has a contract with one of its customers to
fill an order by the end of July.
• There is uncertainty about whether this deadline can be
met, due to uncertainty about whether Bendrix will receive
the materials it needs from one of its suppliers by the
middle of July. It is currently July 1.
• How can the uncertainty in this situation be assessed?
2
SOME KEY TERMS (DEFINITION)
• Random experiment, events, simple events, sample space,
and complement:
– A random experiment is a process of generating
(simple) events. The (simple) events generated by a
random experiment cannot be predicted with certainty
– Simple events cannot be broken down, or decomposed,
into two or more constituent outcomes
– An event is a collection of the simple events of interest.
So, a simple event is also an event.
– The sample space is the list of all possible events
3
SOME KEY TERMS (DEFINITION)
– Complement of an event A is the event that A does not
occur. Sometimes, complement of an event is not just
an event, but a set of many events. So, a more precise
definition is that the complement of an event A is the set
of all events other than A. Complement of an event A is
denoted by AC or A
Event
A
AC
Venn Diagram
Sample
Space
4
SOME KEY TERMS (EXAMPLE)
– Consider the problem of finding a simple random
sample of 10 families from a list 40 families given in file
RANDSAMP.XLS
– The process of finding a simple random sample is a
random experiment
• The process generated a simple random sample.
• A (simple) event is a sample consisting of families
1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Another (simple)
event is a sample consisting of families 1, 2, 3, 4, 5,
6, 7, 8, 9 and 11.
5
SOME KEY TERMS (EXAMPLE)
• There is a large number of ways (how many?) we
can choose 10 families from a list of 40 families.
The sample space consists of all possible
combination of 10 families e.g.,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
1, 2, 3, 4, 5, 6, 7, 8, 9, 11
1, 2, 3, 4, 5, 6, 7, 8, 9, 12
…
6
SOME KEY TERMS (EXAMPLE)
• Example of an event: A sample of 10 families that
includes 5 families with above average income and
5 families with below average income
– this event is a subset of sample space
– this event is a collection of events
– notice that a simple event is an even, but an
event is not necessarily a simple event
7
SOME KEY TERMS (EXAMPLE)
• Let (simple) event A be the sample consisting of
families 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. It’s
complement is the set of all possible combination of
10 families except A e.g.,
1, 2, 3, 4, 5, 6, 7, 8, 9, 11
1, 2, 3, 4, 5, 6, 7, 8, 9, 12
1, 2, 3, 4, 5, 6, 7, 8, 9, 13
…
8
PROBABILITY ESSENTIALS
• Concept of probability is quite intuitive; however, the
rules of probability are not always intuitive or easy to
master.
• Mathematically, a probability is a number between 0 and
1 that measures the likelihood that some event will occur.
– An event with probability zero cannot occur.
– An event with probability 1 is certain to occur.
– An event with probability greater than 0 and less than
1 involves uncertainty, but the closer its probability is
to 1 the more likely it is to occur.
9
Supplier
meets due
date
BENDRIX
COMPANY’S
SITUATION
• Bendrix
collects their
records on
the same
supplier and
similar
contracts
and the data
is shown on
right:
Bendrix
meets due
date
Event B
Supplier
does not
meet due
date
30
4
34
10
16
26
40
20
60
Total
Event BC
Event A
Bendrix
does not
meet due
date
Event AC
Total
10
PROBABILITY ESSENTIALS
• Consider the Bendrix Company data.
• Compute the likelihood that the due date of the contract
will be met. Probability(A occurs), P(A)
• Compute the likelihood that the supplier will meet the due
date. Probability(B occurs), P(B)
11
PROBABILITY ESSENTIALS
• Consider the Bendrix Company data.
• Compute the likelihood that the due date of the contract
will be made if the supplier meets the due date.
Probability(A occurs given that B has occurred), P(A|B)
12
PROBABILITY ESSENTIALS
• Consider the Bendrix Company data.
• Compute the likelihood that the due date of the contract
will be made even if the supplier fails to meet the due
date. Probability(A occurs given that BC has occurred),
P(A|BC)
13
PROBABILITY ESSENTIALS
• Consider the Bendrix Company data.
• Compute the likelihood that both the contract due date
and supplier due date will be met. Probability(A and B
both occur), P(A and B)
14
PROBABILITY ESSENTIALS
• Consider the Bendrix Company data.
• Compute the likelihood that either the contract due date
or the supplier due date will be met. Probability(A occurs
or B occurs or both occur), P(A or B)
15
PROBABILITY RULES AND TREES
•
•
•
•
Rule of complement
Addition rule
Multiplication rule
Probability tree
16
RULE OF COMPLEMENT
• The simplest probability rule involves the
complement of an event.
• If the probability of A is P(A), then the probability of
its complement, P(Ac), is
P(Ac)=1- P(A)
• Equivalently, the probability of an event and the
probability of its complement sum to 1.
P(A) + P(Ac)=1
17
RULE OF COMPLEMENT
THE BENDRIX SITUATION
• Find P(Bc) using the rule of complements
• Does the rule of complements give the same result as it is
given by the frequencies?
Event
B
BC
Venn Diagram
Sample
Space
18
ADDITION RULE
MUTUALLY EXCLUSIVE EVENTS
• We say that events are mutually exclusive if at most one
of them can occur. That is, if one of them occurs, then none
of the others can occur.
• Let A1 through An be any n mutually exclusive events. Then
the addition rule of probability involves the probability that at
least one of these events will occur.
P(at least one of A1 through An) = P(A1) + P(A2) +  + P(An)
19
ADDITION RULE
EXHAUSTIVE EVENTS
• Events can also be exhaustive, which means that they
exhaust all possibilities. Probabilities of exhaustive events
add up to 1.
• If A and B are exhaustive,
P(A)+ P(B)=1
• If A, B and C are exhaustive,
P(A)+ P(B)+ P(C)=1
20
ADDITION RULE
THE BENDRIX SITUATION
• Interpret the events
E1 = (A and B)
E2 = (A and BC)
Sample
Space
A
B
Venn Diagram
21
ADDITION RULE
THE BENDRIX SITUATION
• Are the events E1 and E2 mutually exclusive?
• Verify the following
P(A) = P(E1)+P(E2)
Sample
Space
A
B
Venn Diagram
22
ADDITION RULE
THE BENDRIX SITUATION
• Find P(A) using the relationship P(A) = P(E1)+P(E2), if the
relationship is correct
• Are the events
E1 and E2 exhaustive?
Sample
Space
A
B
Venn Diagram
23
MULTIPLICATION RULE
INDEPENDENT EVENTS
• We say that two events are independent if occurrence
of one does not change the likeliness of occurrence of
the other
• If A and B are two independent events, the joint
probability P(A and B) is obtained by the
multiplication rule.
P(A and B) = P(A)P(B)
24
CONDITIONAL PROBABILITY
• Probabilities are always assessed relative to the information
currently available. As new information becomes available,
probabilities often change.
• A formal way to revise probabilities on the basis of new
information is to use conditional probabilities.
• Let A and B be any events with probabilities P(A) and P(B).
Typically the probability P(A) is assessed without
knowledge of whether B does or does not occur. However if
we are told B has occurred, the probability of A might
change.
25
CONDITIONAL PROBABILITY
• The new probability of A is called the conditional
probability of A given B. It is denoted P(A|B).
– Note that there is uncertainty involving the event to the
left of the vertical bar in this notation; we do not know
whether it will occur or not. However, there is no
uncertainty involving the event to the right of the
vertical bar; we know that it has occurred.
• The following formula conditional probability formula
enables us to calculate P(A|B):
P( A | B) 
P( A and B)
P( B)
26
CONDITIONAL PROBABILITY
• If A and B are two mutually exclusive events, at most one
of them can occur. So,
P(A|B) =0
P(B|A) =0
• If A and B are two independent events, occurrence of one
does not change the likeliness of occurrence of the other.
So,
P(A|B) = P(A)
P(B|A) = P(B)
27
MULTIPLICATION RULE
FOR ANY TWO EVENTS
• In the conditional probability rule the numerator is the
probability that both A and B occur. It must be known in
order to determine P(A|B).
• However, in some applications P(A|B) and P(B) are known;
in these cases we can multiply both side of the conditional
probability formula by P(B) to obtain the multiplication
rule.
P(A and B) = P(A|B)P(B)
• The conditional probability formula and the multiplication
rule are both valid; in fact, they are equivalent.
28
MULTIPLICATION RULE
THE BENDRIX SITUATION
• Are the events A and B independent?
• Find P(A and B) using the multiplication rule
• Does the multiplication
rule give the same result
as it is given by the
A
frequencies?
Sample
Space
B
Venn Diagram
29
PROBABILITY TREES
• Probability trees are useful to
– calculate probabilities
– identify all simple events
– visualize the relationship among the events
• Probability trees are useful if it is possible to
– break down simple events into stages
– identify mutually exclusive and exhaustive events at
each stage
– ascertain the probabilities of events at each stage
30
PROBABILITY TREES
• A probability tree consists of some nodes and branches
• Nodes
– an initial unlabelled node called origin
– other nodes, each labeled with the event represented
by the node
31
PROBABILITY TREES
• Branches
– each branch connect a pair of nodes.
– a branch from A to B implies that event B may occur
after event A
– each branch from
• origin to A is labeled with probability P(A)
• A to B is labeled with the probability P(B|A)
32
PROBABILITY TREES
• Any path through the tree from the origin to a terminal
node corresponds to one possible simple event.
• All simple events and their probabilities are shown next to
the terminal nodes.
33
PROBABILITY TREES
• Example 1: Construct a probability tree diagram for the
Bendrix Company.
Stage 1
P(
B)
=2
/3
B
Stage 2
=3/4
P(A|B)
P(A C|B
)=1/4
C
B
P(
C )=1/5
A
BA
P(BA)=0.5000
AC
BAC
P(BAC)=0.1667
A
BCA
P(BCA)=0.0667
AC
BCAC
P(BCAC)=0.0266
P(A|B
3
1/
)=
BC
Simple Joint
Events Probabilities
P(A C|B C
)=4/5
34
PROBABILITY TREES
• Example 2: In a bag containing 7 red chips and 5 blue
chips you select 2 chips one after the other without
replacement. Construct a probability tree diagram for this
information.
Stage 1
Stage 2
=6/11
)
R
|
R
(
P
P(
R
)=
7/
12
R
P(B|R)
=5
/
=5
B)
P(
=7/11
P(R|B)
/11
B
Simple Joint
Events Probabilities
RR
P(RR)=7/22
B
RB
P(RB)=35/132
R
BR
P(BR)=35/132
B
BB
P(BB)=5/33
12
R
P(B|B)
=4
/11
35
PROBABILITY TREES
• What is the probability of getting a red chip first and then a
blue chip?
• What is the probability of getting a blue chip first and then
a red chip?
• What is the probability of getting a red and a blue chip?
• What is the probability of getting 2 red chips?
36