Transcript Chapter 3 - Pegasus @ UCF

Chapter 3 Probability
3.1 Terminology
3.2 Assign Probability
3.3 Compound Events
3.4 Conditional Probability
3.5 Rules of Computing Probabilities
3.6 Random Sampling
Homework:3, 7, 17, 23, 27, 29, 32, 35, 37, 43, 50,
55, 57, 61
1
Section 3.1: Some terminology of probabilities
We discussed how to understand/describe the
information contained in a sample in Chapter II.
However, we want to make inference based on the
information contained in a sample as well. We
will discuss the concept of probability in this
chapter because probability plays an important
role in inference making process. Let start our
discussion with one example.
2
<Example 3.1>: (Basic) Suppose there is a bag of
M&M chocolate with six different coating colors -- 300
brown, 250 red, 200 yellow, 150 orange, 100 green, and
100 tan. Suppose one piece is drawn at random and the
coating color is recorded.
(a) Is this a random experiment?
(b) How many sample points in this random experiment?
(c) What is the sample space of this random experiment?
(d) What is the probability of drawing a yellow M&M
chocolate?
(e) What is the event of drawing a piece of M&M with your
favorite colors if your favorite colors are yellow and
green?
3
Section 3.2: Assign probability to a sample point and to
an event
There are two approaches, the relative frequency
approach and subjective approach, to assign a probability
to a sample point. Both approaches need to follow two
basic rules. The first rule is that the probability for each
sample point must lie between 0 and 1. The second rule
is that the probabilities of all the sample points within a
sample space must sum to 1. We use the subjective
approach to assign probability in Example 3.2 and 3.4
because the frequency tables are unavailable, and use the
relative frequency approach to assign probability in
Example 3.1 and 3.3 because the frequency tables are
available. Usually, we use relative frequency approach to
assign probability if the frequency table is available.
4
We can employ the following five steps to compute the
probability of an event.
(1) Define the experiment.
(2) List all the sample points.
(3) Assign probability to each sample point.
(4) Find out all the sample points in this event.
(5). Sum the probabilities of these sample points.
5
<Example 3.2>: (Basic)
Two fair coins are randomly tossed, and their up faces are
recorded.
(a) Is this a random experiment?
(b) Write down the sample space.
(c) Assign probabilities to each sample points.
(d) Event A is at least one head. Event B is at least one tail.
Compute the probability of event A and event B.
6
<Example 3.3>: (Basic)
Suppose that we repeat toss a pair of fair coins ten
thousand times. Table 3.1 is the frequency table of this
random experiment.
Table 3.1
Simple Event
TT
TH
HT
HH
Frequencies
1977 2855 2423 2745
(a) Assign probabilities to all simple events.
(b) Event A is at least one head. Event B is at least one tail.
Compute the probabilities of event A and event B.
7
<Example 3.4>: (Basic)
Two fair dice are tossed, and the up face on each die is
recorded.
(a) Write down the sample space.
(b) Assign probabilities to all the sample points.
(c)A = {sum of two numbers is equal to 7.}
B = {sum of two numbers is greater than or equal to 8.}
C = {3 on the up face of at least one die.}
Find the probabilities of events A, B, and C.
8
Section 3.3 Compound Events, Complementary
Events, and Mutually Exclusiveness
A compound event is a composition of two or
more events. A compound event can be formed
through either union or intersection operation.
The union of two events A and B denoted by the
symbol AB consists all the sample points that
belong to A or B or both. The intersection of two
events A and B denoted by the symbol AB
consists all the sample points belonging to both A
and B.
9
<Example 3.5>: (Basic) Continuation of Example 3.4.
Event D = {Sum of two numbers is equal to 10}
Event E ={Difference of two numbers is equal to 2}
Event F = {Two even numbers}
Event G = {Two odd number}
Event H = {Sum of two numbers is larger than or equal to
10}
(a) Find the probabilities of events D, E, F, G, and H.
(b) Find the probability of D  E.
(c) Find the probability of D  E.
10
<Solutions>:
(a)Event D = {(4,6), (5,5), (6,4)}
Event E = {(1,3), (2,4), (3,5), (4,6), (6,4), (5,3), (4,2), (3,1)}
Event F = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)}
Event G = {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}
Event H = {(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}
P(D) = 3/36 = 1/12;P(E) = 8/36 = 2/9;P(F) = 9/36 = 1/4;
P(G) = 9/36 = 1/4; P(H) = 6/36 = 1/6.
(b) Event D  E = {(4,6), (6,4)} and P(D  E) = 2/36 = 1/18.
(c) Event D  E = {(1,3), (2,4), (3,5), (4,6), (5,5), (6,4), (5,3),
(4,2), (3,1)} and P(D 
11
The complement of a event A, denoted by Ac, is the
event that A does not occur, i.e. Ac consists all the
sample points not belonging to event A. The
concept of complementary event is very important
in computing the event probability because that in
many probability problems calculating the
probability of the complement of the event of
interest is easier than calculating the probability of
the event itself.
12
<Example 3.6> (Basic) Continuation of Example
3.4
(a) Is Event F a complementary event of Event G?
(b) Find the probability of Fc , Gc , and Hc.
13
Event A and Event B are mutually exclusive if A B
contains no sample points, that is, A and B have no sample
points in common. A and Ac has no sample points in
common, i.e. A and Ac are mutually exclusive.
<Example 3.7> (Basic) Continuation of Example 3.4
(a) Are events F and G mutually exclusive?
(b) Find P(F G).
<Solutions>:
(a) Events F and g have no sample points in common, i.e.
they are mutually exclusive.
(b) P(F  G) = 0.
14
Note:
(a) The sum of the probabilities of complementary
events equals one; that is, P(A) + P(Ac) = 1.
(b) Event A and event Ac have no sample points in
common.
(c) Event A and event Ac are mutually exclusive.
(d) P(A) + P(Ac) = 1. This means the Event A and
the event Ac cover the entire sample space.
(e) Venn Diagram is a good graphical tool for
understanding the concept of compound events,
complementary events, and mutually
exclusiveness.
15
Section 3.4: Conditional Probability
(Intermediate)
The event probabilities we have been discussing based
on subjective approach or relative frequency approach are
unconditional probabilities because no special conditions
are assumed when we compute these probabilities.
However, we sometimes we have additional information
that might alter the probability of a given event. A
probability that reflects such additional knowledge is
called conditional probability of the event. The
probability that event A occurs given that the event B
occurs is called the conditional probability of A given B
and can be denoted by the symbol P(A|B), where P(A|B) =
P(AB)/P(B) if P(B) > 0.
16
<Example 3.8> (Basic)
Consider the experiment of tossing a fair coin twice and
recording the up face on each toss. The following event are
defined:
A: {the first toss is a head}
B: {the second toss is a head}
Find P(A), P(B), P(B|A), and P(BA).
<solutions>:
Sample space = {HH, HT, TH, TT}
A = {HH, HT}; B = {HH, TH}; BA = {HH}
P(A) = 2/4 = ½; P(B) = 2/4 = ½; P(BA) = ¼;
P(B|A) = P(BA)/P(A) = (1/4)/(1/2) = ½.
17
<Example 3.9>: (Basic) Suppose that the sample space
is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, Event A is {2, 5, 6, 7, 0}, and
Event B is {0, 4, 7, 8}. Find P(A), P(B), P(A  B),
P(A|B), P(B|A), P(A  B), and P(A’).
<Solutions>:
A  B  {0, 7}
A  B = {0, 2, 4, 5, 6, 7, 8}
P(A) = 5/10 = 0.5;
P(B) = 4/10 = 0.4;
P(A  B) = 2/10 = 0.2;
P(A|B) = P(A  B  P(B) = 0.2/0.4 = 0.5;
P(B|A) = P(B  A  P(A  0205  0;
P(A  B) = 7 / 10 = 0.7;
P(A’) = 1 - P(A) = 1 - 0.5 = 0.5.
18
<Example 3.10>: (Intermediate)
Five
hundred
married men and women were asked if they would marry
their current spouses if they were given a chance to do it
over again. Their responses are recorded in the following :
Table 3.2
Yes
No
Male
125
175
Female
55
145
(a). The following events are defined:
Event A: Want to marry their current spouses again
Event B: Female; Event C: Male
Event D: Do not want to marry to their current spouses
again;
Find P(A), P(A|B), P(B|A), and P(A  B)
(b). Are the events B and D mutually exclusive? Explain. 19
<Solutions>:
(a) P(A) = (125+55)/500 = 0.36;
P(B) = (55+145)/500 = 0.4;
P(C ) = (125+175)/500 = 0.6;
P(D) = (175+145)/500 = 0.64;
P(A  B  55500  0;
A(A|B  P(A  BP(B  0.11/0.4 = 0.275;
P(B|A) = P(B  AP(A) = 0.11/0.36 = 11/36;
P(A  B) = (125+55+145)/500 = 0.665.
(b)
P(B  D) = 145 / 500 = 0.29. Thus, events B and D
are not mutually exclusive because there are many
unhappy wives who do not want to marry their current
spouses again.
20
Section 3.5: Rules of Computing the Probabilities
Both additive rule and multiplicative rule are
very useful when we want to compute the probabilities of
compound events. The additivity rule of probability is
used to find the probability of union of two events and the
multiplicative rule of probability is used to find the
probability of intersection of two events. Let A and B be
two events, the additivity rule of probability is P(A  B) =
P(A) + P(B) - P(A  B) and the multiplicative rule of
probability is P(A  B) = P(A|B)*P(B) = P(B|A)*P(A).
21
We say two events A and B are independent
events if the probability of the occurrence of one event
(say A) does not alter the probability that another event
(say B) has occurred. P(A|B) = P(A) and P(B|A) = P(B)
when A and B are independent events. If events A and B
are independent, the multiplicative rule of probability
becomes to P(AB) = P(A)*P(B) = P(B)*P(A). We say
events A and B dependent events if A and B are not
independent.
Note:
(1) P(A  B) = P(A) + P(B) if events A and B are mutually
exclusive because P(AB) = 0.
(2) P(A  B) = P(A) * P(B) if events A and B are
independent events.
22
<Example 3.11>: (Basic)
When an American marriage ends in divorce, it is most
likely that the women is the partner who is dissatisfied. A
study of National Center for Health Statistics reported the
following table:
1975
1986
By wife
67.2%
61.5%
By husband
29.4%
32.6%
Jointly
3.4%
5.9%
Suppose that two women are interviewed, one of whom
was divorced in 1975, and the other in 1986.
(a) What is the probability both were filed by themselves?
(b) What is the probability of the first was filed by herself and
the second was filed by her exit-husband?
23
<solutions>:
(a) Let A: {divorced in 1975 and filed by herself}
B: {divorced in 1986 and filed by herself}
C: {divorced in 1986 and filed by her exit-husband}
then P(A) = 0.672, P(B) = 0.615, and P(C) = 0.326.
Since
A
and
B
are
independent
events,
P(AB)=P(A)*P(B)=0.672*0.615=0.41328.
(b) Since A and C are independent events, P(AC) =
P(A)*P(C ) = 0.672*0.326 = 0.219072.
24
<Example 3.12>: (Intermediate)
A smoke detector system uses two devices, P and Q.
If smoke is present, the probability that it will be detected
by device P is 0.95: by device Q is 0.98: by both devices is
0.94. Considering the following events:
A: {The smoke will be detected by device P}
B: {The smoke will be detective by device Q}
(a) Find the probability of smoke will be detected by at least
one device.
(b) Find the probability that the smoke will not be detected.
25
<Solutions>:
(a) P(A) = 0.95, P(B) = 0.98, and P(AB) = 0.94 are given.
Applying the additivity rule of probability, we have P(A 
B) = P(A) + P(B) - P(AB) = 0.95 + 0.98 - 0.94 = 0.99.
(b) Smoke will not be detected is a complement event of the
event that smoke will be detected by at least one device.
Applying the probability of complement, we have P(smoke
will not be detected) = 1 - P(A  B) = 1 - 0.99 = 0.01.
26
<Example 3.13>: (Advance)
In a study involving a manufacturing process, 15% of all
parts tested were defective, and 40% of all parts were
produced by machine I. If a part was produced by machine
I there is a 20% chance that it is defective.
Let A = {the part is produced by machine I}
B = {the part is defective}.
(a) Find P(A) and P(B).
(b) Find P(B|A).
(c) Find P(A  B).
(d) If a part is found defective, what is the probability that it
came from machine I?
(e). Are A and B independent events? Explain.
27
<Solutions>:
(a) P(A) = 40% = 0.4 and P(B) = 15% = 0.15.
(b) P(B|A) = 0.20.
Note: (1) The condition is the part was produced by machine I, i.e., event
A. (2) P(B|A) is the conditional probability that the defective part was
produced by machine I. (3) Thus, P(B|A) is given in the problem and
P(B|A) = 0.20.
(c) P(A  B) = P(B|A) * P(A) = 0.20 * 0.4 = 0.08.
(d) P(A|B) = P(A  B) / P(B) = 0.08 / 0.15 = 8/15.
Note: (1) The condition now is the part was found defective. (2) P(A|B)
is the conditional probability that the part produced by machine is
defective. (3) Thus, P(A|B) = 8/15.
(e) P(A) * P(B) = 0.4 * 0.15 = 0.06 and P(A  B) = 0.08.
Thus, P(A  B) P(A) * P(B) and A and B are not
independent events.
28
Section 3.6 Random Sampling
How a sample is selected from a population is of vital
importance in statistical inference. Although there are
many sampling procedures can be used to obtain a useful
sample in statistical inference, we will only discuss the
most frequently and simplest form of sampling procedure,
simple random sampling procedure, in this semester. A
sample of size n is called a simple random sample (or
random sample) if every set of n elements in the
population has equal chance of being selected. Usually, we
rely on random number generator that are built into
most statistical software to generate the random sample.
We say a sample is biased if not all set of n elements has
equal chance of being selected. Usually, we want to avoid
to use a biased sample in statistical inference.
29
Collection of Definitions:
Random Experiment A random experiment is a
process that can be repeated under the same conditions.
And each outcome of this process can not be predicted in
advance without uncertainty.
Sample Point
A sample point is the most basic
outcomes from a random experiment.
Sample Space Sample space is the collection of all
possible sample points from a random experiment.
30
Event An event is a specific collection of sample points.
Probability of a sample point The probability of a
sample point is a number between 0 and 1 that reflects the
chance that the outcome will occur when the experiment is
performed.
Union The union of two events A and B is the event that
either A or B or both occur in a single trail of the
experiment. We denote the union of A and B by the
symbol A  B.
31
Intersection The intersection of A and B is the event
that both A and B occur n a single trail of the experiment.
We denote the intersection of A and B by the symbol A B.
Complementary Events The complement of an
event A is the event that A does not occur. This means the
complement event of A consists all sample points that are
not in event A. We denote the complement of event by the
symbol A’ or AC.
Mutually Exclusive Events
Events A and B are
mutually exclusive events if event A and event B have no
sample points in common.
32
Conditional Probability
The probability that event A occurs given that the condition
of event B occurs is called the conditional probability of A
given B. The conditional probability of event A given
event B is denoted by the symbol P(A|B), and P(A|B) =
P(AB)/P(B) if P(B) > 0.
Additive Rule of Probability
The probability of the union of events A and B, denoted by
P(AB), equals P(A) + P(B) - P(AB). In particular, P(A 
B) = P(A) + P(B) if events A and B are mutually exclusive.
33
Multiplicative Rule of Probability
The probability of the intersection of events A and B,
denoted by P(AB), equals to P(A)*P(B|A) or equals to
P(B) * P(A|B). P(AB) = P(A) * P(B) if events A and B are
independent.
Independent and Dependent
Events A and B are independent events if the occurrence of
B does not alter the probability that A has occurred. That
is P(A|B) = P(A) and P(B|A) = P(B). Events A and B are
dependent if they are not independent,
34