Download Chapter 3 Probability

Chapter 3
Probability
Probability is the tool that allows the statistician to use sample information to make inferences about or to describe the population from which the sample was drawn.
3.1
Events, Sample Spaces, and Probability
Definition 3.1 Experiment: An experiment is the process by which an observation (or
measurement) is obtained.
Definition 3.2 Sample Point: A sample point is the most basic outcome of an experiment.
Definition 3.3 Sample Space: The sample space is the set of all possible outcomes (sample
points) of an experiment and is denoted by the symbol S.
Example 3.1, page 118: Two coins are tossed and their up faces are recorded. List of all
sample points for this experiment. Find the sample space S.
See examples for experiments and sample spaces in Table 3.1 on page 119.
Extra Example 1: Toss a fair coin and through a die. List of all sample points for this
experiment and find the sample space S.
Venn Diagram: A pictorial method for presenting the sample space where each sample
point is represented by a solid dot and labeled accordingly is called the Venn diagram. See
Figure 3.2, page 120.
Probability: The probability of a sample point is a number between 0 and 1 that measures
the likelihood that the outcome will occur when the experiment is performed.
Probability Rules for Sample Points:
Let pi represent the probability of a sample point i. Then
1. All sample point probabilities must lie between 0 and 1 (0 ≤ pi ≤ 1).
2. The probabilities of all the sample points within a sample space must sum to 1 ( i pi = 1).
Example 3.2, page 121.
Example 3.3, page 122.
17
Definition 3.4 Event: An event is a specific collection of sample points.
Simple Event: A single outcome of an experiment is called a simple event.
Compound Event: A compound event is a collection of simple events.
Probability of an event: The probability of an event A is equal to the sum of the probabilities of the simple events in A.
Example 3.4, page 123.
Steps for calculating probabilities of events
1.
2.
3.
4.
5.
Define the experiment and the type of observation that will be recorded.
List the sample points.
Assign probabilities to the sample points.
Determine the collection of sample points contained in the event of interest.
Sum the sample point probabilities to get the event probability.
Extra Example 2: Toss a die and observe the number appearing on the upper face is an
experiment. The sample space is
S = {1, 2, 3, 4, 5, 6}.
We define the following events
Event A = {Observe a odd number}, that is A={1, 3, 5}
Event B= {Observe a number less than 4}, that is B={1, 2, 3}
Event C = {Observe an even number}, that is C={2, 4, 6}
Find the P(A), P(B) and P(C).
Combinations Rule: A sample of n elements is to be drawn from
a set
of N elements.
N
Then the number of different possible samples is denoted by CnN =
and is equal to
n
CnN =
N
n
=
N!
n!(N − n)!
where
n! = n(n − 1)(n − 2) . . . 4.3.2.1 (Note 0! = 1)
Example 3.7, page 126.
Extra Example 3: The personnel director of a company plans to hire two salespeople
from a total of four applicants. Suppose she is completely incapable of correctly ranking the
applicants according to their ability and in effect, selects them at random.
(a) What is the probability that she selects the two best candidates?
(b) What is the probability that she selects at least one of the two best candidates?
18
3.2
Unions and Intersections
Definition 3.5 Union of A and B: Let A and B be two events in a sample space S. The
union of A and B is the event containing all simple events in A or B or both. We denote the
union of A and B by the symbol A ∪ B.
Definition 3.6 Intersection of A and B: Let A and B be two events in a sample space
S. The intersection of A and B is the event composed of all simple events that are in both
A and B. We denote the intersection of A and B by the symbol A ∩ B or simply AB.
Extra Example 4: An experiment can result in 1 of 10 simple events E1 , E2 , . . . E10 which
are equally likely; the events A, B, and C are defined as follows
Event
Simple events
A
E1 E2 E3 E4
B
E3 E4 E5 E6 E7
C
E6 E7 E8
(a) List the simple events in the following compound events A∪B, AB, AC, B ∪C, A∪B ∪C
and A ∩ B ∩ C.
(b) Calculate the probabilities associated with each of the events in part (a) by summing
the probabilities of the appropriate simple events.
Example 3.9, page 131.
3.3
Complementary Events
Definition 3.7 Complementary event: The complement of an event A, denoted by Ac,
consists of all the simple events in the sample space that are not in A. For a complementary
event A,
P (Ac ) = 1 − P (A).
The above equation implies
P (Ac ) + P (A) = 1.
Example 3.11, page 134.
3.4
The Additive Rule and Mutually Exclusive Events
Additive Rule of Probability: The probability of the union of events A and B is the sum
of probabilities of events A and B minus the probability of the intersection of events A and
B. Given two events A and B, the probability that A or B or both occur is
P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
19
Example 3.12, page 135.
Definition 3.8 Mutually Exclusive Event: Two events A and B are said to be mutually
exclusive (ME) if when A occurs, B can not occured (and vice versa). That means, two
events A and B are said to be mutually exclusive if the event A ∩ B contains no simple
event. That is if P (A ∩ B) = 0. Therefore, the events A and Ā are mutually exclusive.
If two events A and B are mutually exclusive, then
P (A ∪ B) = P (A) + P (B).
Example 3.13, page 136.
Exercise 3.34, page 138.
3.5
Conditional Probability
Sometimes we have additional knowledge/information that might affect the likelihood of
the outcome of an experiment or that can be used to better determine the probability of
the event of interest. A probability that reflects such additional knowledge is called the
conditional probability.
If A and B are any two events, then the conditional probability of A given B, denoted by
P (A|B) is
P (A ∩ B)
P (A|B) =
, P (B) > 0.
P (B)
If A and B are any two events, then the conditional probability of B given A, denoted by
P (B|A) is
P (A ∩ B)
, P (A) > 0.
P (B|A) =
P (A)
Example 3.15, page 146: Application of Conditional Probability, Statistics by
McClave and Sincich
Many medical researchers have conducted experiments to examine the relationship between
smoking cigarette and lung cancer. Let A represent the event that the individual smokes
and let Ac denote the complement of A (the event that the individual does not smoke).
Similarly, let B represent the event that the individual develops cancer and let B c be the
complement of that event. Then the four sample points associated with the experiment are
shown in Table 3.1 and their probabilities for a certain section (a particular population) of
the United Sates are provided in Table 3.2.
20
Table 3.1: Sample space for the above example
A ∩ B A ∩ Bc
Ac ∩ B
Ac ∩ B c
Table 3.2: Probabilities of smoking and
Develops Cancer
Smoker Yes, B No, B c
Yes, A
0.05
0.20
No, Ac
0.03
0.72
Total
0.08
0.92
developing cancer
Total
0.25
0.75
1.00
Suppose an individual has randomly selected from this population.
(a) What is the probability that the randomly selected individual is a smoker?
Ans:
P (A) = P (A ∩ B) + P (A ∩ B c ) = 0.05 + 0.20 = 0.25
(b) What is the probability that a randomly selected individual develops cancer? Ans:
P (B) = P (A ∩ B) + P (Ac ∩ B) = 0.05 + 0.03 = 0.08
(c) What is the probability that a randomly selected smoker develops cancer? OR
What is the probability that a randomly selected individual develops cancer given that he/she
smokes cigarette? OR
If you know an individual smoker in that area, what is the probability that he/she develops
cancer?
Ans:
P (B|A) =
0.05
P (A ∩ B)
=
= 0.20
P (A)
0.25
(d) What is the probability that a randomly selected smoker does not develop cancer? OR
What is the probability that a randomly selected individual does not develop cancer given
that he/she smokes cigarette? OR
If you know an individual smoker in that area, what is the probability that he/she does not
develop cancer?
Ans:
P (B c |A) =
P (A ∩ B c )
0.20
=
= 0.80
P (A)
0.25
21
(e) What is the probability that a randomly selected individual develops cancer given that
he/she does not smoke cigarette?
Ans:
P (Ac ∩ B)
0.03
P (B|Ac) =
=
= 0.04
c
P (A )
0.75
(f ) What is the probability that a randomly selected individual does not develop cancer
given that he/she does not smoke cigarette? Ans:
P (B c |Ac ) =
P (Ac ∩ B c )
0.72
=
= 0.96
c
P (A )
0.75
The conditional probability that an individual smoker develops cancer (0.20) is five times
the probability that a nonsmoker develops cancer (0.04). This does not necessarily imply
that smoking causes cancer, but it does suggest a link between smoking and cancer.
3.6
The Multiplicative Rule and Independent Events
Multiplicative Rule of Probability: The probability that both A and B occur is
P (A ∩ B) = P (AB) = P (A|B)P (B)
= P (B|A)P (A).
(3.1)
Example 3.17, page 145.
Definition 3.9 Independent Events: Two events A and B are said to be independent if
any one of the followings holds:
(i) P (A|B) = P (A)
(ii) P (B|A) = P (B)
(iii) P (A ∩ B) = P (A)P (B).
Otherwise, the events are said to be dependent.
Example 3.19, page 147.
Note: Mutually exclusive does not necessarily mean independent. For example, the events
A and C in extra example 2 are mutually exclusive but they are not independent.
Probability of Intersection of Two Independent Events
If events A and B are independent, then
P (A ∩ B) = P (A)P (B)
The converse is also true: If P (A)P (B) = P (A ∩ B), then events A and B are independent.
Exercise 3.50, page 151.
22
3.7
Random Sampling
Definition 3.10 Simple Random Sample: Let N and n represent the number of elements
in the population and sample respectively. If the sampling is conducted in such a way that
!
samples has an equal probability of being selected, the sampling
each of the CnN = n!(NN−n)!
is said to be random and the resulting sample is said to be a simple random sample.
Here
N!
CnN =
,
n!(N − n)!
where n! = n(n − 1)(n − 2)......4.3.2.1. Then 5! = 5.4.3.2.1 = 120.
Example 3.22, page 155.
23