Probability

1
Chapter 5
5-1

• Probability

2
Outline
5-2

• 5-1 Introduction
• 5-2 Sample Spaces and Probability
• 5-3 The Addition Rules for
  Probability
• 5-4 The Multiplication Rules and Conditional
  Probability

3
Objectives
5-3

• Determine Sample Spaces and find the probability
  of an event using classical probability.
• Find the probability of an event using empirical
  probability.
• Find the probability of compound events using the
  addition rules.

4
Objectives
5-4

• Find the probability of compound events using the
  multiplication rules.
• Find the conditional probability of an event.

5
5-2 Sample Spaces and Probability
5-5

• A probability experiment is a process that leads
  to well-defined results called outcomes.
• An outcome is the result of a single trial of a
  probability experiment.
• NOTE A tree diagram can be used as a systematic
  way to find all possible outcomes of a
  probability experiment.

6
5-2 Tree Diagram for Tossing Two Coins
5-6

First Toss
7
5-2 Sample Spaces - Examples
5-7
8
5-2 Formula for Classical Probability
5-8

• Classical probability assumes that all outcomes
  in the sample space are equally likely to occur.
  
• That is, equally likely events are events that
  have the same probability of occurring.

9
5-2 Formula for Classical Probability
5-9
10
5-2 Classical Probability - Examples
5-10

• For a card drawn from an ordinary deck, find the
  probability of getting (a) a queen (b) a 6 of
  clubs (c) a 3 or a diamond.
• Solution (a) Since there are 4 queens and 52
  cards, P(queen) 4/52 1/13.
• (b) Since there is only one 6 of clubs, then P(6
  of clubs) 1/52.

11
5-2 Classical Probability - Examples
5-11

• (c) There are four 3s and 13 diamonds, but the 3
  of diamonds is counted twice in the listing.
  Hence there are only 16 possibilities of drawing
  a 3 or a diamond, thus P(3 or
  diamond) 16/52 4/13.

12
5-2 Classical Probability - Examples
5-12

• When a single die is rolled, find the probability
  of getting a 9.
• Solution Since the sample space is 1, 2, 3, 4,
  5, and 6, it is impossible to get a 9. Hence,
  P(9) 0/6 0.
• NOTE The sum of the probabilities of all
  outcomes in a sample space is one.

13
5-2 Complement of an Event
5-13
E
14
5-2 Complement of an Event - Example
5-14

• Find the complement of each event.
• Rolling a die and getting a 4.
• Solution Getting a 1, 2, 3, 5, or 6.
• Selecting a letter of the alphabet and getting a
  vowel.
• Solution Getting a consonant (assume y is a
  consonant).

15
5-2 Complement of an Event - Example
5-15

• Selecting a day of the week and getting a
  weekday.
• Solution Getting Saturday or Sunday.
• Selecting a one-child family and getting a boy.
• Solution Getting a girl.

16
5-2 Rule for Complementary Event
5-16
?
?
P
E
P
E
(
)
(
)
1

or
P
E
P
E
1
(
)

?
(
)
or
P
E
P
E
1
(
)

(
)

.
17
5-2 Empirical Probability
5-17

• The difference between classical and empirical
  probability is that classical probability assumes
  that certain outcomes are equally likely while
  empirical probability relies on actual experience
  to determine the probability of an outcome.

18
5-2 Formula for Empirical Probability
5-18
19
5-2 Empirical Probability - Example
5-19

• In a sample of 50 people, 21 had type O blood, 22
  had type A blood, 5 had type B blood, and 2 had
  AB blood. Set up a frequency distribution.

20
5-2 Empirical Probability - Example
5-20
Type
Frequency
A B AB O
22 5 2 21
50 n
21
5-2 Empirical Probability - Example
5-21

• Find the following probabilities for the previous
  example.
• A person has type O blood.
• Solution P(O) f /n 21/50.
• A person has type A or type B blood.
• Solution P(A or B) 22/50 5/50
  27/50.

22
5-3 The Addition Rules for Probability
5-22

• Two events are mutually exclusive if they cannot
  occur at the same time (i.e. they have no
  outcomes in common).

23
5-3 The Addition Rules for Probability
5-23
A and B are mutually exclusive
A
B
24
5-3 Addition Rule 1
5-24





When two events A and B are mutually exclusive,
the probabilitythat A or B will occur is












25
5-3 Addition Rule 1- Example
5-25

• At a political rally, there are 20 Republicans
  (R), 13 Democrats (D), and 6 Independents (I).
  If a person is selected, find the probability
  that he or she is either a Democrat or an
  Independent.
• Solution P(D or I) P(D) P(I)
  13/39 6/39 19/39.

26
5-3 Addition Rule 1- Example
5-26

• A day of the week is selected at random. Find
  the probability that it is a weekend.
• Solution P(Saturday or Sunday)
  P(Saturday) P(Sunday)
  1/7 1/7 2/7.

27
5-3 Addition Rule 2
5-27
When two events A and B






are not mutually exclusive, the





probability
y
that
A
or
B
will






occur
is


?
?
?
P
A
or
B
P
A
P
B
P
A
and
B





(
)
(
)
(
)
(
)
28
5-3 Addition Rule 2
5-28
A and B
(common portion)
A
B
29
5-3 Addition Rule 2- Example
5-29

• In a hospital unit there are eight nurses and
  five physicians. Seven nurses and three
  physicians are females. If a staff person is
  selected, find the probability that the subject
  is a nurse or a male.
• The next slide has the data.

30
5-3 Addition Rule 2 - Example
5-30
31
5-3 Addition Rule 2 - Example
5-31

• Solution P(nurse or male)
  P(nurse) P(male) P(male nurse) 8/13 3/13
  1/13 10/13.

32
5-3 Addition Rule 2 - Example
5-32

• On New Years Eve, the probability that a person
  driving while intoxicated is 0.32, the
  probability of a person having a driving accident
  is 0.09, and the probability of a person having a
  driving accident while intoxicated is 0.06. What
  is the probability of a person driving while
  intoxicated or having a driving accident?

33
5-3 Addition Rule 2 - Example
5-33

• Solution P(intoxicated or
  accident)
  P(intoxicated) P(accident)
  P(intoxicated and accident)
  0.32 0.09 0.06 0.35.

34
5-4 The Multiplication Rules and
Conditional Probability
5-34

• Two events A and B are independent if the fact
  that A occurs does not affect the probability of
  B occurring.
• Example Rolling a die and getting a 6, and then
  rolling another die and getting a 3 are
  independent events.

35
5-4 Multiplication Rule 1
5-35
36
5-4 Multiplication Rule 1 - Example
5-36

• A card is drawn from a deck and replaced then a
  second card is drawn. Find the probability of
  getting a queen and then an ace.
• Solution Because these two events are
  independent (why?), P(queen and ace)
  (4/52)?(4/52) 16/2704 1/169.

37
5-4 Multiplication Rule 1 - Example
5-37

• A Harris pole found that 46 of Americans say
  they suffer great stress at least once a week.
  If three people are selected at random, find the
  probability that all three will say that they
  suffer stress at least once a week.
• Solution Let S denote stress. Then P(S and S
  and S) (0.46)3 0.097.

38
5-4 Multiplication Rule 1 - Example
5-38

• The probability that a specific medical test will
  show positive is 0.32. If four people are
  tested, find the probability that all four will
  show positive.
• Solution Let T denote a positive test result.
  Then P(T and T and T and T) (0.32)4 0.010.

39
5-4 The Multiplication Rules and
Conditional Probability
5-39

• When the outcome or occurrence of the first event
  affects the outcome or occurrence of the second
  event in such a way that the probability is
  changed, the events are said to be dependent.
• Example Having high grades and getting a
  scholarship are dependent events.

40
5-4 The Multiplication Rules and
Conditional Probability
5-40

• The conditional probability of an event B in
  relationship to an event A is the probability
  that an event B occurs after event A has already
  occurred.
• The notation for the conditional probability of B
  given A is P(BA).
• NOTE This does not mean B ? A.

41
5-4 Multiplication Rule 2
5-41
42
5-4 The Multiplication Rules and
Conditional Probability - Example
5-42

• In a shipment of 25 microwave ovens, two are
  defective. If two ovens are randomly selected
  and tested, find the probability that both are
  defective if the first one is not replaced after
  it has been tested.
• Solution See next slide.

43
5-4 The Multiplication Rules and
Conditional Probability - Example
5-43

• Solution Since the events are dependent, P(D1
  and D2) P(D1)?P(D2 D1)
  (2/25)(1/24) 2/600 1/300.

44
5-4 The Multiplication Rules and
Conditional Probability - Example
5-44

• The WW Insurance Company found that 53 of the
  residents of a city had homeowners insurance
  with its company. Of these clients, 27 also had
  automobile insurance with the company. If a
  resident is selected at random, find the
  probability that the resident has both
  homeowners and automobile insurance.

45
5-4 The Multiplication Rules and
Conditional Probability - Example
5-45

• Solution Since the events are dependent, P(H and
  A) P(H)?P(AH)
  (0.53)(0.27) 0.1431.

46
5-4 The Multiplication Rules and
Conditional Probability - Example
5-46

• Box 1 contains two red balls and one blue ball.
  Box 2 contains three blue balls and one red ball.
  A coin is tossed. If it falls heads up, box 1
  is selected and a ball is drawn. If it falls
  tails up, box 2 is selected and a ball is drawn.
  Find the probability of selecting a red ball.

47
5-4 Tree Diagram for Example
5-47

48
5-4 The Multiplication Rules and
Conditional Probability - Example
5-48

• Solution P(red) (1/2)(2/3) (1/2)(1/4) 2/6
  1/8 8/24 3/24 11/24.

49
5-4 Conditional Probability - Formula
5-49
50
5-4 Conditional Probability - Example
5-50

• The probability that Sam parks in a no-parking
  zone and gets a parking ticket is 0.06, and the
  probability that Sam cannot find a legal parking
  space and has to park in the no-parking zone is
  0.2. On Tuesday, Sam arrives at school and has
  to park in a no-parking zone. Find the
  probability that he will get a ticket.

51
5-4 Conditional Probability - Example
5-51

• Solution Let N parking in a no-parking zone
  and T getting a ticket.
• Then P(T N) P(N and T) /P(N) 0.06/0.2
  0.30.

52
5-4 Conditional Probability - Example
5-52

• A recent survey asked 100 people if they thought
  women in the armed forces should be permitted to
  participate in combat. The results are shown in
  the table on the next slide.

53
5-4 Conditional Probability - Example
5-53
54
5-4 Conditional Probability - Example
5-54

• Find the probability that the respondent answered
  yes given that the respondent was a female.
• Solution Let M respondent was a male F
  respondent was a female Y
  respondent answered yes N
  respondent answered no.

55
5-4 Conditional Probability - Example
5-55

• P(YF) P( F and Y) /P(F) 8/100/50/100
  4/25.
• Find the probability that the respondent was a
  male, given that the respondent answered no.
• Solution P(MN) P(N and M)/P(N)
  18/100/60/100 3/10.