Download Statistics Chapter 5

Definitions
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
1
Notation
P - denotes a probability
A, B, ... - denote specific events
P (E) -
denotes the probability of event E
occurring
2
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
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 A actually occurs, then an estimate of P(E) is
P(E) =
Number of times E occurred
Total number of possible outcomes
4
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
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
number of ways E can occur
= number of different simple
=
events
P(E)
s
n
6
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.
7
Rule 1
The relative frequency approach is an
approximation.
8
Rule 1
The relative frequency approach is an
approximation.
Rule 2
The classical approach is the actual
probability.
9
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.
10
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?
11
Example:
Find the probability that a randomly
selected person will be struck by lightning this year.
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
12
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
13
Probability Limits
 The probability of an impossible event is 0.
 The probability of an event that is certain to
occur is 1.
14
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
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
Impossible
to occur
Certain
to occur
16
Possible Values for Probabilities
1
Certain
Likely
0.5
50-50 Chance
Unlikely
0
Impossible
17
Complementary Events
18
Complementary Events
c
E,
The complement of event E, denoted by
consists of all outcomes in which event E
does not occur.
19
Complementary Events
c
E,
The complement of event E, denoted by
consists of all outcomes in which event E
does not occur.
P(E)
C
P(E )
(read “not E”)
20
Example:
Testing Corvettes
The 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?
21
Example:
Testing Corvettes
The 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?
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
22
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
23
A Tree Diagram for Tossing a Coin
Twice
There are 4 possible outcomes when tossing a
coin twice.
First Toss
Second Toss
Outcomes
H
HH
T
H
HT
TH
T
TT
H
T
24
Rules of Complementary Events
P(A) +
c
P(A)
=1
25
Rules of Complementary Events
P(A) +
c
P(A)
c
P(A)
=1
= 1 - P(A)
26
Possible outcomes for two rolls of a die
(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)
27
Find the following probabilities
1. Find the probability that the sum is a 2
2. Find the probability that the sum is a 3
3. Find the probability that the sum is a 4
4. Find the probability that the sum is a 5
5. Find the probability that the sum is a 6
6. Find the probability that the sum is a 7
7. Find the probability that the sum is a 8
8. Find the probability that the sum is a 9
9. Find the probability that the sum is a 10
10. Find the probability that the sum is a 11
11. 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
28
29
Rounding Off Probabilities
give the exact fraction or decimal
or
30
Rounding Off Probabilities
give the exact fraction or decimal
or
round off the final result to three
significant digits
31
Tree Diagram of Test Answers
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?
32
Tree Diagram of Test Answers
a
b
T
c
d
e
a
b
c
F
d
e
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
33
Tree Diagram of Test Answers
What is the probability that the first
question is true and the second
question is c?
34
Tree Diagram of Test Answers
a
b
T
c
d
e
a
b
c
F
d
e
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
35
FIGURE 3-9
Tree Diagram of Test Answers
a
b
T
c
d
e
a
b
c
F
P(T) =
1
2
d
e
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
36
Tree Diagram of Test Answers
a
b
T
c
d
e
a
b
c
F
P(T) =
1
2
d
e
P(c) =
1
5
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
37
FIGURE 3-9
Tree Diagram of Test Answers
a
b
c
d
T
e
a
b
c
F
P(T) =
1
2
d
e
P(c) =
1
5
Ta
Tb
Tc
Td
Te
Fa
Fb
Fc
Fd
Fe
P(T and c) =
1
10
38
P (both correct) = P (T and c)
1 1 1
 
10 2 5
39
Rock – Paper – Scissors
Tree Diagram- 2 Players
R
R
T
P
P
B
S
A
R
A
S
P
T
S
B
R
B
P
A
S
T
3 1
3 1
3 1
P ( A)  
P( B)  
P (T )  
9 3
9 3
9 3
40
Rock – Paper – Scissors
Tree Diagram- 3 Players
R
R
P
P
S
R
P
S
S
R
P
S
RPSRP SRP SRPSRP SRPSRPSRP SRPS
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
3
1
18
2
6
2
P ( A) 

P( B) 

P (C ) 

27 9
27
3
27 9
41
Pg 189 #9
1/16
5/8
13/14
1/3
3/4
1/3
42
Definition
 Compound Event
•
Any event combining 2
or more simple events
43
44
Definition
 Notation
•
P(A or B) = P (event A
occurs or event B
occurs or they both occur)
45
Compound Event
• 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.
46
Compound Event
• 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.
P(A  B)  P(A)  P(B)  P(A  B)
47
Definition
• Events A and B are mutually exclusive if
they cannot occur simultaneously.
48
Venn Diagrams
Total Area = 1
P(A)
P(B)
Total Area = 1
P(A)
P(B)
P(A and B)
Overlapping Events
Non-overlapping Events
49
Venn Diagrams
50
A  (B  C)
(A  B)  C
(AB)  (A  C)
Ac  B
A  (B C)
U - Ac
(A  B )c
A-B
Ac  Bc  Cc
51
Ac  (B  C)c
Ac  (Bc  C)
(A  B)c
(A B) (A C)
A - (B C)
A  (B  Ac)
(A  B)  (A C)
B-A
(Ac  Bc)  Cc
52
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
20
13
20
2
30
6
4
13C
U
B
20
5
20
P(A B C)  P(A)  P(B)  P(C)  P(A B)  P(B C)  P(A C)  P(A B C)
P(A U B U C) = 42 + 31 + 28 – 6 – 10 -9 + 4 = 80
53
A survey of 500 television watchers produced the following information:
285 watch football
190 watch hockey
A
B
115 watch basketball
25
45 watch football and basketball
45
195
70 watch football and hockey
25
20
45
50 watch hockey and basketball
50 do not watch any sports.
95
a. How many watch all three games?
b. How many watch exactly one of the
three games?
P(A
B C)  P(A)  P(B)  P(C)  P(A
C
U
B)  P(B C)  P(A C)  P(A
50
B C)
500 = 285 + 190 + 115 - 45 - 70 - 50 + P(A  B  C) + 50
500 = 475 + P(A  B  C)
P(A  B  C) = 25
54
Applying the Addition Rule
P(A or B)
Addition Rule
Are
A and B
mutually
exclusive
?
Yes
P(A or B) = P(A) + P(B)
No
P(A or B) = P(A)+ P(B) - P(A and B)
55
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
• Find the probability of randomly
selecting a man or a boy.
56
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
• Find the probability of randomly
selecting a man or a boy.
57
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
• Find the probability of randomly selecting
a man or a boy.
• P(man or boy) =
1692
64
1756


 .790
2223 2223 2223
* Mutually Exclusive *
58
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
• Find the probability of randomly
selecting a man or someone who
survived.
59
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
• Find the probability of randomly
selecting a man or someone who
survived.
60
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
• Find the probability of randomly selecting a man or
someone who survived.
• P(man or survivor) =
1692 706
332 2066



2223 2223 2223 2223
= 0.929
* NOT Mutually Exclusive *
61
Complementary Events
P(A) and
c
P(A)
are mutually exclusive
All simple events are either
c
in A or A
62
Venn Diagram for the
Complement of Event A
Total Area = 1
P (A)
P (A)c = 1 - P (A)
63
Finding the Probability of Two or
More Selections
 Multiple selections
 Multiplication Rule
64
Notation
P(A and B) =
P(event A occurs in a first trial and
event B occurs in a second trial)
65
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.
66
Conditional Probability
P(A and B) = P(A) • P(B|A)
P(B|A) =
P(A and B)
P(A)
P(A  B)

P(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.
67
Notation for Conditional
Probability
 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”).
68
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.
69
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)
70
Applying the Multiplication Rule
P(A or B)
Multiplication Rule
Are
A and B
independent
?
Yes
P(A and B) = P(A) • P(B)
No
P(A and B) = P(A) • P(B|A)
71
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 where
c
P(A)
c
P(A)
is P(getting none)
72
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
c
P(A)
c
P(A)
is P(getting no girls)
= (0.5)(0.5)(0.5) = 0.125
P(A) = 1 - 0.125 = 0.875
73
Testing for Independence
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.
74
Testing for Independence
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)
75
Contingency Table
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Totals
706
1517
2223
Find the probability of randomly selecting a man if
you know the person is a survivor
P(man  survivor)
P(man | survivor) 
P(survivor )
332

 .470
706
Find the probability of selecting a survivor if you know the
person is not a boy
677
c
P(b  survivor)
P(survivor | not a boy) 
P(b c )

2159
 .309
76
Calculate the following probabilities:
Eas t B1
Sou th B2
Midwes t B3
Fa rwe st B4
City Typ e
La rge A1
35
10
25
25
Sma ll A2
15
25
15
15
Sub urb A3
25
5
10
10
75
a. P(A1) 95/215= 19/43
50/215=10/43
b. P(B3)
c. P (A1  B4) 25/215 = 5/43
25/50=1/2
d. P(B1|A3)
40
e. P(A2 U B3)
f. P(B1 U B4)
g. P( B2 B4)
h. P(A2|B4)
50
50
95
70
50
215
105/215=21/43
125/215=25/43
none
15/50 = 3/10
77
Find :
a. P(A B)
P(A) = 1/3
P(B) = 1/4
P(A U B) = 1/2
a.
1/12
b. P(A | B)
b.
1/3
c. P(B | A)
c.
1/4
d.
1/4
e.
1/2
f.
2/3
g.
2/3
h.
11/12
c
A
3/12
B
1/12
2/12
d. P(A B )
e. P(A
c
c
B)
c
f. P(A | B)
c
c
g. P(A | B )
6/12
h. P(A
c
c
B)
78