Chapter 3 Probability

1

• 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
• Homework3, 7, 17, 23, 27, 29, 32, 35, 37, 43,
  50,
• 55, 57, 61

2

• 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.

3

• ltExample 3.1gt (Basic) Suppose there is a bag of
  MM 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
  MM chocolate?
• (e) What is the event of drawing a piece of MM
  with your favorite colors if your favorite colors
  are yellow and green?

4

• 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.

5

• 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.

6

• ltExample 3.2gt (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.

7

• ltExample 3.3gt (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.

8

• ltExample 3.4gt (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.

9

• 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.

10

• ltExample 3.5gt (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 Ç E.
• (c) Find the probability of D È E.

11

• ltSolutionsgt
• (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 1/12P(E) 8/36 2/9P(F) 9/36
  1/4
• P(G) 9/36 1/4 P(H) 6/36 1/6.
• (b) Event D Ç E (4,6), (6,4) and P(D Ç E)
  2/36 1/18.
• (c) Event D È E (1,3), (2,4), (3,5), (4,6),
  (5,5), (6,4), (5,3), (4,2), (3,1) and P(D È E)
  9/36 1/4.

12

• 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.

13

• ltExample 3.6gt (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.

14

• Event A and Event B are mutually exclusive if
  AÇ 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.
• ltExample 3.7gt (Basic) Continuation of Example 3.4
• (a) Are events F and G mutually exclusive?
• (b) Find P(FÇ G).
• ltSolutionsgt
• (a) Events F and g have no sample points in
  common, i.e. they are mutually exclusive.
• (b) P(F Ç G) 0.

15

• 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.

16

• 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(AB), where
  P(AB) P(AB)/P(B) if P(B) gt 0.

17

• ltExample 3.8gt (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(BA), and P(BA).
• ltsolutionsgt
• Sample space HH, HT, TH, TT
• A HH, HT B HH, TH BÇA HH
• P(A) 2/4 ½ P(B) 2/4 ½ P(BÇA) ¼
• P(BA) P(BÇA)/P(A) (1/4)/(1/2) ½.

18

• ltExample 3.9gt (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 Ç B), P(AB), P(BA), P(A È
  B), and P(A).
• ltSolutionsgt
• A Ç B 0, 7
• A È B 0, 2, 4, 5, 6, 7, 8
• P(A) 5/10 0.5
• P(B) 4/10 0.4
• P(A Ç B) 2/10 0.2
• P(AB) P(A Ç B) / P(B) 0.2/0.4 0.5
• P(BA) P(B Ç A) / P(A) 0.2/0.5 0.4
• P(A È B) 7 / 10 0.7
• P(A) 1 - P(A) 1 - 0.5 0.5.

19

• ltExample 3.10gt (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(AB), P(BA), and P(A È B)
• (b). Are the events B and D mutually exclusive?
  Explain.

20

• ltSolutionsgt
• (a) P(A) (12555)/500 0.36
• P(B) (55145)/500 0.4
• P(C ) (125175)/500 0.6
• P(D) (175145)/500 0.64
• P(A Ç B) 55/500 0.11
• A(AB) P(A Ç B)/P(B) 0.11/0.4 0.275
• P(BA) P(B Ç A)/P(A) 0.11/0.36 11/36
• P(A È B) (12555145)/500 0.665.
• (b) P(B Ç 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.

21

• 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 È B) P(A)
  P(B) - P(A Ç B) and the multiplicative rule of
  probability is P(A Ç B) P(AB)P(B)
  P(BA)P(A).

22

• 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(AB)
  P(A) and P(BA) 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 È B) P(A) P(B) if events A and B are
  mutually exclusive because P(AB) 0.
• (2) P(A Ç B) P(A) P(B) if events A and B are
  independent events.

23

• ltExample 3.11gt (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?

24

• ltsolutionsgt
• (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(AÇB)P(A)P(B)0.6720.6150.41328.
• (b) Since A and C are independent events, P(AÇC)
  P(A)P(C ) 0.6720.326 0.219072.

25

• ltExample 3.12gt (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.

26

• ltSolutionsgt
• (a) P(A) 0.95, P(B) 0.98, and P(AÇB) 0.94
  are given. Applying the additivity rule of
  probability, we have P(A È B) P(A) P(B) -
  P(AÇB) 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 È B) 1 - 0.99 0.01.

27

• ltExample 3.13gt (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(BA).
• (c) Find P(A Ç 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.

28

• ltSolutionsgt
• (a) P(A) 40 0.4 and P(B) 15 0.15.
• (b) P(BA) 0.20.
• Note (1) The condition is the part was produced
  by machine I, i.e., event A. (2) P(BA) is the
  conditional probability that the defective part
  was produced by machine I. (3) Thus, P(BA) is
  given in the problem and P(BA) 0.20.
• (c) P(A Ç B) P(BA) P(A) 0.20 0.4 0.08.
• (d) P(AB) P(A Ç B) / P(B) 0.08 / 0.15
  8/15.
• Note (1) The condition now is the part was found
  defective. (2) P(AB) is the conditional
  probability that the part produced by machine is
  defective. (3) Thus, P(AB) 8/15.
• (e) P(A) P(B) 0.4 0.15 0.06 and P(A Ç B)
  0.08. Thus, P(A Ç B) P(A) P(B) and A and B
  are not independent events.

29

• 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.

30

• 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.

31

• 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.

32

• 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 .
• Mutually Exclusive Events Events A and B
  are mutually exclusive events if event A and
  event B have no sample points in common.

33

• 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(AB), and P(AB)
  P(AB)/P(B) if P(B) gt 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.

34

• Multiplicative Rule of Probability
• The probability of the intersection of events
  A and B, denoted by P(AB), equals to P(A)P(BA)
  or equals to P(B) P(AB). 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(AB) P(A) and
  P(BA) P(B). Events A and B are dependent if
  they are not independent,