Date post: | 05-Jan-2016 |
Category: |
Documents |
Upload: | beau-shepherd |
View: | 28 times |
Download: | 0 times |
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
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 A actually occurs, then an estimate of P(E) is
P(E) = Number of times E occurredTotal number of possible outcomes
5
Basic Rules for Computing Probability
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
Basic Rules for Computing Probability
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
Basic Rules for Computing Probability
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.
11
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?
12
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
14
Probability Limits
The probability of an impossible event is 0.
The probability of an event that is certain to occur is 1.
15
Probability Limits
The probability of an impossible event is 0.
The probability of an event that is certain to occur is 1.
• 0 P(A) 1
16
Probability Limits
The probability of an impossible event is 0.
The probability of an event that is certain to occur is 1.
0 P(A) 1
Impossibleto occur
Certainto occur
17
Possible Values for ProbabilitiesCertain
Likely
50-50 Chance
Unlikely
Impossible
1
0.5
0
18
Complementary Events
19
Complementary Events
The complement of event E, denoted by Ec, consists of all outcomes in which event E
does not occur.
20
P(E)
Complementary Events
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
24
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
25
Rules of Complementary Events
P(A) + P(A)c = 1
26
P(A)c
Rules of Complementary Events
P(A) + P(A)c = 1
= 1 - P(A)
27
(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
28
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
29
30
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
32
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
33
TaTbTcTdTeFaFbFcFdFe
a
b
c
d
e
a
b
c
d
e
T
F
Tree Diagram of Test Answers
34
What is the probability that the first question is true and the second
question is c?
Tree Diagram of Test Answers
35
TaTbTcTdTeFaFbFcFdFe
a
b
c
d
e
a
b
c
d
e
T
F
Tree Diagram of Test Answers
36
TaTbTcTdTeFaFbFcFdFe
a
b
c
d
e
a
b
c
d
e
T
F
P(T) =
FIGURE 3-9 Tree Diagram of Test Answers
12
37
TaTbTcTdTeFaFbFcFdFe
a
b
c
d
e
a
b
c
d
e
T
F
P(T) = P(c) =
Tree Diagram of Test Answers
12
15
38
TaTbTcTdTeFaFbFcFdFe
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
43
Compound Event• Any event combining 2 or more simple events
Definition
44
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
47
• 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
48
Definition• Events A and B are mutually exclusive if
they cannot occur simultaneously.
49
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
51
A (B C) Ac B (A B )c
A (B C)(A B) C A - B
(AB) (A C) U - Ac Ac Bc Cc
52
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
53
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
54
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
55
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
56
• 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
• 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
58
• Find the probability of randomly selecting a man or a boy.
• P(man or 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
.7902223
1756
2223
64
2223
1692
* Mutually Exclusive *
59
• Find the probability of randomly selecting a man or someone who survived.
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
60
• Find the probability of randomly selecting a man or someone who survived.
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
61
• Find the probability of randomly selecting a man or someone who survived.
• P(man or survivor) =
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
* NOT Mutually Exclusive *
= 0.9292223
2066
2223
332
2223
706
2223
1692
62
Complementary Events
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.
67
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
P(A and B) = P(A) • P(B|A)
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|A)
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
74
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
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
)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
78
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