1 event - any collection of results or outcomes from some procedure simple event - any outcome or...

1

Event - any collection of results or outcomes from some procedure

Simple event - any outcome or event that cannot be broken down into simpler components

Compound event – an event made up of two or more other events

Sample space - all possible simple events

Definitions

2

Notation P - denotes a probability

A, B, ... - denote specific events

P (E) - denotes the probability of event E occurring

3

Basic Rules for Computing Probability

Rule 1: Relative Frequency Approximation

Conduct (or observe) an experiment a large number of times, and count the number of times event E actually occurs, then an estimate of P(E) is

4


Rule 1: Relative Frequency Approximation

Conduct (or observe) an experiment a large number of times, and count the number of times event A actually occurs, then an estimate of P(E) is

P(E) = Number of times E occurredTotal number of possible outcomes

5


Rule 2: Classical approach (requires equally likely outcomes)

If a procedure has n different simple events, each with an equal chance of occurring, and s is the number of ways event E can occur, then

6


Rule 2: Classical approach (requires equally likely outcomes)

If a procedure has n different simple events, each with an equal chance of occurring, and s is the number of ways event E can occur, then

P(E) = number of ways E can occurnumber of different simple

events

sn =

7


Rule 3: Subjective Probabilities

P(E), the probability of E, is found by simply guessing or

estimating its value based on knowledge of the relevant

circumstances.

8

Rule 1 The relative frequency approach is an

approximation.

9

Rule 1 The relative frequency approach is an

approximation.

Rule 2 The classical approach is the actual

probability.

10

Law of Large Numbers

As a procedure is repeated again and again, the relative frequency probability (from Rule 1) of an event tends to approach the actual probability.

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Law of Large Numbers

Flip a coin 20 times and record the number of heads after each trial. In L1 list the numbers 1-20, in L2 record the number of heads.In L3, divide L2 by L1. Get a scatter plot with L1 and L3. What can you conclude?

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The sample space consists of two simple events: the person is struck by lightning or is not. Because these simple events are not equally likely, we can use the relative frequency approximation (Rule 1) or subjectively estimate the probability (Rule 3). Using Rule 1, we can research past events to determine that in a recent year 377 people were struck by lightning in the US, which has a population of about 274,037,295. Therefore,

P(struck by lightning in a year)

377 / 274,037,295 1/727,000

Example: Find the probability that a randomly selected person will be struck by lightning this year.

13

Example: On an ACT or SAT test, a typical multiple-choice question has 5 possible answers. If you make a random guess

on one such question, what is the probability that your response is wrong?

There are 5 possible outcomes or answers, and there are 4 ways to answer incorrectly. Random guessing implies that the outcomes in the sample space are equally likely, so we apply the classical approach (Rule 2) to get:

P(wrong answer) = 4 / 5 = 0.8

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Probability Limits

The probability of an impossible event is 0.

The probability of an event that is certain to occur is 1.

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Probability Limits



• 0 P(A) 1

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Probability Limits



0 P(A) 1

Impossibleto occur

Certainto occur

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Possible Values for ProbabilitiesCertain

Likely

50-50 Chance

Unlikely

Impossible

1

0.5

0

18

Complementary Events

19


The complement of event E, denoted by Ec, consists of all outcomes in which event E

does not occur.

20

P(E)


The complement of event E, denoted by Ec, consists of all outcomes in which event E does not occur.

P(EC)(read “not E”)

21

Example: Testing CorvettesThe General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 50 drivers has been recruited, 20 or whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver?

22

Because 20 of the 50 subjects are men, it follows that 30 of the 50 subjects are women so,

P(not selecting a man) = P(man)c

= P(woman) = 30 = 0.6

50

Example: Testing CorvettesThe General Motors Corporation wants to conduct a test of a new model of Corvette. A pool of 50 drivers has been recruited, 20 or whom are men. When the first person is selected from this pool, what is the probability of not getting a male driver?

23

Using a Tree Diagram

Flipping a coin is an experiment and the possible outcomes are heads (H) or tails (T).

One way to picture the outcomes of an experiment is to draw a tree diagram. Each outcome is shown on a separate branch. For example, the outcomes of flipping a coin are

H

T

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A Tree Diagram for Tossing a Coin Twice

There are 4 possible outcomes when tossing a coin twice.

H

TH

T

H

T

First Toss Second Toss Outcomes

HH

HTTH

TT

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Rules of Complementary Events

P(A) + P(A)c = 1

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P(A)c

Rules of Complementary Events

P(A) + P(A)c = 1

= 1 - P(A)

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(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Possible outcomes for two rolls of a die

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1. Find the probability that the sum is a 22. Find the probability that the sum is a 33. Find the probability that the sum is a 44. Find the probability that the sum is a 55. Find the probability that the sum is a 66. Find the probability that the sum is a 77. Find the probability that the sum is a 88. Find the probability that the sum is a 99. Find the probability that the sum is a 1010. Find the probability that the sum is a 1111. Find the probability that the sum is a 12

• 1/36• 2/36• 3/36• 4/36• 5/36• 6/36• 5/36• 4/26• 3/36• 2/36• 1/36

Find the following probabilities

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Rounding Off Probabilities

give the exact fraction or decimal

or

31

Rounding Off Probabilities

give the exact fraction or decimal

orround off the final result to three significant digits

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How many ways are there to answer a two question test when the first question is a true-false question and the second question is a multiple choice question with five possible answers?

Tree Diagram of Test Answers

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TaTbTcTdTeFaFbFcFdFe

a

b

c

d

e

a

b

c

d

e

T

F


34

What is the probability that the first question is true and the second

question is c?


35


a

b

c

d

e

a

b

c

d

e

T

F


36


a

b

c

d

e

a

b

c

d

e

T

F

P(T) =

FIGURE 3-9 Tree Diagram of Test Answers

12

37


a

b

c

d

e

a

b

c

d

e

T

F

P(T) = P(c) =


12

15

38


a

b

c

d

e

a

b

c

d

e

T

F

P(T) = P(c) = P(T and c) =

FIGURE 3-9 Tree Diagram of Test Answers

12

15

110

39

P (both correct) = P (T and c)

5

1

2

1

10

1

40

R P S

R P S R P S R P S

Rock – Paper – Scissors Tree Diagram- 2 Players

T B A A T B B A T

3

1

9

3)(

3

1

9

3)(

3

1

9

3)( TPBPAP

41

R P S

R P S R P S R P S

RPSRP SRP SRPSRPSRPSRP SRPSRPS

Rock – Paper – Scissors Tree Diagram- 3 Players

A B B B B C B C B B B C B A B C B B B C B C B B B B A

9

2

27

6)(

3

2

27

18)(

9

1

27

3)( CPBPAP

42

1/165/8

13/14

1/3

3/4

1/3

Pg 189 #9

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Compound Event• Any event combining 2 or more simple events

Definition

45

Notation

• P(A or B) = P (event A occurs or event B occurs or they both occur)

Definition

46

• General Rule

When finding the probability that event A occurs or

event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find the total in such a way that no outcome is counted more than once.

Compound Event

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• Formal Addition Rule• P(A or B) = P(A) + P(B) - P(A and B)

• where P(A and B) denotes the probability that A and B both occur at the same time.

Compound Event

B)P(AP(B)P(A)B)P(A

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Definition• Events A and B are mutually exclusive if

they cannot occur simultaneously.

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Venn Diagrams

Total Area = 1

P(A) P(B)

Non-overlapping Events

P(A) P(B)

P(A and B)

Overlapping Events

Total Area = 1

50Venn Diagrams

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A (B C) Ac B (A B )c

A (B C)(A B) C A - B

(AB) (A C) U - Ac Ac Bc Cc

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Ac (Bc C)Ac (B C)c (A B)c

(A B) (A C) A - (B C) A (B Ac)

(A B) (A C) B - A (Ac Bc) Cc

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A poll was taken of 100 students to find out how they arrived at school. 28 used car pools; 31 used buses; and 42 said they drove to school alone.In addition, 9 used both car pools and buses;10 used car pools and drove alone; only 6 used buses and their own car and 4 used all three methods.

a. Complete the Venn diagram.b. How many used none of the methods?c. How many used only car pools?d. How many used buses exclusively?

A B

UC

46 5

230 20

1320

201320

P(ABC) P(A) P(B) P(C) P(AB) P(BC) P(AC) P(ABC)

P(A U B U C) = 42 + 31 + 28 – 6 – 10 -9 + 4 = 80

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A survey of 500 television watchers produced the following information:285 watch football190 watch hockey115 watch basketball45 watch football and basketball70 watch football and hockey50 watch hockey and basketball50 do not watch any sports.

a. How many watch all three games?b. How many watch exactly one of the three games?

A B

UC

P(ABC) P(A) P(B) P(C) P(AB) P(BC) P(AC) P(ABC)

500 = 285 + 190 + 115 - 45 - 70 - 50 + P(A B C) + 50

500 = 475 + P(A B C) P(A B C) = 25

2545 20

45195

95

25

50

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Applying the Addition Rule

P(A or B)

Addition Rule

AreA and Bmutuallyexclusive

?

P(A or B) = P(A)+ P(B) - P(A and B)

P(A or B) = P(A) + P(B)Yes

No

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• Find the probability of randomly selecting a man or a boy.

Men Women Boys Girls Totals

Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

57



Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

58


• P(man or boy) =


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

.7902223

1756

2223

64

2223

1692

* Mutually Exclusive *

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• Find the probability of randomly selecting a man or someone who survived.


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

60



Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

61


• P(man or survivor) =


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

* NOT Mutually Exclusive *

= 0.9292223

2066

2223

332

2223

706

2223

1692

62


P(A) and P(A)c are mutually exclusive

All simple events are either in A or Ac

63

Venn Diagram for the Complement of Event A

Total Area = 1

P (A)

P (A)c = 1 - P (A)

64

Finding the Probability of Two or More Selections

Multiple selections

Multiplication Rule

65

NotationP(A and B) =

P(event A occurs in a first trial and

event B occurs in a second trial)

66

Conditional Probability

Definition

The conditional probability of event B occurring, given that A has already occurred, can be found by dividing the probability of events A and B both occurring by the probability of event A.

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Conditional Probability

P(A and B) = P(A) • P(B|A)

The conditional probability of B given A can be found by assuming the event A has occurred and, operating under that assumption, calculating the probability that event B will occur.

P(A)

B)P(A

P(A and B)P(A)P(B|A) =

68

P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B|A as “B given A”).

Notation for Conditional Probability

69

Definitions

Independent Events• Two events A and B are independent if the occurrence of

one does not affect the probability of the occurrence of the other.

Dependent Events• If A and B are not independent, they are said to be

dependent.

70

Formal Multiplication Rule


If A and B are independent events, P(B|A) is really the same as P(B)

71

Applying the Multiplication Rule

P(A or B)

Multiplication Rule

AreA and B

independent?


P(A and B) = P(A) • P(B)Yes

No

72

Probability of ‘At Least One’ ‘At least one’ is equivalent to ‘one or more’.

The complement of getting at least one item of a particular type is that you get no items of that type.

If P(A) = P(getting at least one), then

P(A) = 1 - P(A)c

where P(A)c is P(getting none)

73

Probability of ‘At Least One’ Find the probability of a couple have

at least 1 girl among 3 children. If P(A) = P(getting at least 1 girl), then

P(A) = 1 - P(A)c

where P(A)c is P(getting no girls)

P(A)c = (0.5)(0.5)(0.5) = 0.125

P(A) = 1 - 0.125 = 0.875

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If P(B|A) = P(B)

then the occurrence of A has no effect on the probability of event B; that is, A and B are independent events.

Testing for Independence

75

If P(B|A) = P(B)

then the occurrence of A has no effect on the probability of event B; that is, A and B are independent events.

or

If P(A and B) = P(A) • P(B)

then A and B are independent events.

Testing for Independence

76

Find the probability of randomly selecting a man if you know the person is a survivor


Survived 332 318 29 27 706

Died 1360 104 35 18 1517

Total 1692 422 64 45 2223

Contingency Table

)P(survivor

survivor)P(mansurvivor)|P(man

.470

706

332

)P(b

survivor)P(bboy)anot |P(survivor

c

c .309

2159

677

Find the probability of selecting a survivor if you know the person is not a boy

77

Calculate the following probabilities:

a. P(A1) e. P(A2 U B3) b. P(B3) f. P(B1 U B4)c. P (A1 B4) g. P( B2 B4)d. P(B1|A3) h. P(A2|B4)

East B1 South B2 Midwest B3 Farwest B4

City Type

Large A1 35 10 25 25

Small A2 15 25 15 15

Suburb A3 25 5 10 10

75 40 50 50

957050

215

95/215= 19/43

50/215=10/43

25/215 = 5/43

25/50=1/2

105/215=21/43

125/215=25/43

none

15/50 = 3/10

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P(A) = 1/3P(B) = 1/4P(A U B) = 1/2

Find :

a. P(AB)

b. P(A | B)

c. P(B | A)

d. P(ABc)

e. P(Ac Bc)

f. P(Ac | B)

g. P(Ac | Bc)

h. P(Ac Bc)

1/123/12 2/12

6/12

A B

a. 1/12

b. 1/3

c. 1/4

d. 1/4

e. 1/2

f. 2/3

g. 2/3

h. 11/12

1 event - any collection of results or outcomes from some procedure simple event - any outcome or...

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