Transcript Chap. 3: Probability - Department of Statistics and Probability

Statistics for Business and
Economics
Chapter 3
Probability
Contents
1.
2.
3.
4.
Events, Sample Spaces, and Probability
Unions and Intersections
Complementary Events
The Additive Rule and Mutually Exclusive
Events
Learning Objectives
1. Develop probability as a measure of
uncertainty
2. Introduce basic rules for finding
probabilities
3. Use probability as a measure of reliability
for an inference
4. Provide an advanced rule for finding
probabilities
Thinking Challenge
• What’s the probability
of getting a head on
the toss of a single fair
coin? Use a scale from
0 (no way) to 1 (sure
thing).
• So toss a coin twice.
Do it! Did you get one
head & one tail?
What’s it all mean?
Many Repetitions!*
Total Heads
Number of Tosses
1.00
0.75
0.50
0.25
0.00
0
25
50
75
Number of Tosses
100
125
3.1
Events, Sample Spaces,
and Probability
Experiments & Sample Spaces
1. Experiment
• Process of observation that leads to a single
outcome that cannot be predicted with certainty
2. Sample point
• Most basic outcome of an
experiment
Sample Space
Depends on
Experimenter!
3. Sample space (S)
• Collection of all sample points
Visualizing
Sample Space
1.
Listing for the experiment of tossing a coin once and
noting up face
S = {Head, Tail}
Sample point
2.
A pictorial method for presenting the sample space
Venn Diagram
H
T
S
Example
• Experiment: Tossing two coins and
recording up faces:
• Is sample space as below?
S={HH, HT, TT}
Tree Diagram
1st coin
H
T
2nd coin
H
T
H
T
Sample Space Examples
•
•
•
•
•
•
•
Experiment
Sample Space
Toss a Coin, Note Face
Toss 2 Coins, Note Faces
Select 1 Card, Note Kind
Select 1 Card, Note Color
Play a Football Game
Inspect a Part, Note Quality
Observe Gender
{Head, Tail}
{HH, HT, TH, TT}
{2♥, 2♠, ..., A♦} (52)
{Red, Black}
{Win, Lose, Tie}
{Defective, Good}
{Male, Female}
Events
1. Specific collection of sample points
2. Simple Event
• Contains only one sample point
3. Compound Event
• Contains two or more sample points
Venn Diagram
Experiment: Toss 2 Coins. Note Faces.
Sample Space
Outcome;
Sample
point
S = {HH, HT, TH, TT}
TH
HH
Compound
Event: At
least one
Tail
HT
TT
S
Venn Diagram
Experiment: Toss 2 Coins. Note Faces.
Sample Space
S = {HH, HT, TH, TT}
TH
TT
HH
Compound
Event:
Exactly one
head
HT
S
Venn Diagram
Experiment: Toss 2 Coins. Note Faces.
Sample Space
S = {HH, HT, TH, TT}
Compound
Event: Tail
at the 2nd
toss
TH
HT
HH
TT
S
Venn Diagram
Experiment: Toss 2 Coins. Note Faces.
Sample Space
S = {HH, HT, TH, TT}
HT
TH
TT
HH
S
Simple
Event: Tail
for both
tosses
Event Examples
Experiment: Toss 2 Coins. Note Faces.
Sample Space: HH, HT, TH, TT
•
•
•
•
Event
1 Head & 1 Tail
Head on 1st Coin
At Least 1 Head
Heads on Both
Outcomes in Event
HT, TH
HH, HT
HH, HT, TH
HH
Thinking challenge
• A fair coin is tossed till to get the first head or
four tails in a row. Which one is the sample
space for this experiment?
a. S={T, TH, TTH, TTTH, TTTT}
b. S={T, HT, TTH, TTTH, TTTT}
c. S={H, TH, TTH, TTTH, TTTT}
d. S={H, HT, HHT, HHHT, HHHH}
Probabilities
What is Probability?
1. Numerical measure of the
likelihood that event will
occur
• P(Event)
• P(A)
• Prob(A)
1
Certain
.5
2. Lies between 0 & 1
3. Sum of probabilities for all
0
sample points in the
sample space is 1
Impossible
Probability Rules
for Sample Points
Let pi represent the probability of sample point i.
1. All sample point probabilities must lie between 0
and 1 (i.e., 0 ≤ pi ≤ 1).
2. The probabilities of all sample points within a
sample space must sum to 1 (i.e.,  pi = 1).
Equally Likely Probability
P(Event) = X / T
• X = Number of outcomes in the
event
• T = Total number of sample points
in Sample Space
• Each of T sample points is equally
likely
— P(sample point) = 1/T
© 1984-1994 T/Maker Co.
Steps for Calculating Probability
1. Define the experiment; describe the process used to
make an observation and the type of observation
that will be recorded
2. List the sample points
3. Assign probabilities to the sample points
4. Determine the collection of sample points contained
in the event of interest
5. Sum the sample points probabilities to get the event
probability
Thinking Challenge
• Consider the experiment of tossing two balanced dice.
Which of the following are true for this experiment.
I. The probability of having 4 or less for the sum of the dots
on the up faces is 1/6.
II. The probability of having sum of the dots on the upfaces
larger than 4 is 5/6
III.The probability of having 6 or less for the sum of the dots
on the up faces is 7/12
a. I and II
b. I and III
c. II and III d. all
Thinking Challenge (sol.)
• Sample space for tossing two dice
S={(1,1),(1,2),…(1,6),(2,1),…,(2,6),…,(6,6)}
with 36 sample points
• Let event A=Having 4 or less for the sum of
upfaces. So,
• A={(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)}
• Prob:{1/36,1/36,1/36,1/36,1/36,1/36}
• P(A)=6/36=1/6
Thinking Challenge (sol.)
• Let event B=Having the sum of upfaces
larger than 4. So,
• B={(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,
2),(3,3),…(3,6),(4,1),…(4,6),(5,1),…(5,6),(6,
1),…(6,6)}
With 30 sample points each with prob.1/36
• P(B)=30/36=5/6
Thinking Challenge (sol.)
• Let event C=Having the sum of upfaces 6 or
less. So,
• C={ (1,1), (1,2), (1,3), (1,4), (1,5), (2,1),
(2,2), (2,3), (2,4), (3,1), (3,2),(3,3), (4,1),
(4,2), (5,1)}
each with prob.1/36
• P(C)=15/36=5/12
So the correct choice is a. I and II
Combinations Rule
A sample of n elements is to be drawn from a set of N
elements. The, the number of different samples possible
 N
is denoted by   and is equal to
n 
 N
N!
 n   n!N  n !
where the factorial symbol (!) means that
n!=n*(n-1)*…*3*2*1
For example, 5!  5  4  3 2 1
0! is defined to be 1.
Thinking Challenge
• The price of a european tour includes four
stopovers to be selected from among 10
cities. In how many different ways can one
plan such a tour if the order of the
stopovers does not matter?
Ex.3.2 from text book (p.139)
P(A)=0.3, P(B)=0.2
P(A)=0.25, P(B)=0.3
Ex.3.9 from text book
Ex.3.9 from text book: solution
a. S={Brown, yellow, red, blue, orange,green}
b. P={0.13, 0.14, 0.13, 0.24, 0.20, 0.16}
c. Let event A=selecting brown candy
P(A)=P(Brown)=0.13
d. Let event B=selecting red, green or yellow candy
P(B)= 0.13+0.16+0.14=0.43
e. Let event C= selecing a candy other than blue
P(C) = 0.13 +0.14+ 0.13+ 0.20+ 0.16=0.76
or P(C) = 1-0.24=0.76
Ex.3.25 from text book
Ex.3.25 from text book: solution
• Odds in favor of E= P(E) / [1-P (E)]
• Odds against E = [1-P (E)] / P(E)
a. Odds in favor of Oxford Shoes
winning=(1/3)/(2/3)=1/2 meaning 1 to 2
b. 1/1= P(E) / [1-P (E)]
P(E)=1/2
c. Odds against Oxford Shoes winning = 3/2
3/2 = [1-P (E)] / P(E)
P(E)=2/5
3.2
Unions and Intersections
Compound Events
Compound events:
Composition of two or more other events.
Can be formed in two different ways.
Unions & Intersections
1. Union
•
•
•
Outcomes in either events A or B or both
‘OR’ statement
Denoted by  symbol (i.e., A  B)
2. Intersection
•
•
•
Outcomes in both events A and B
‘AND’ statement
Denoted by  symbol (i.e., A  B)
Event Union:
Venn Diagram
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Sample
Space:
2, 2, 2,
..., A
Ace
Event Ace:
A, A, A, A
Black
S
Event Ace  Black:
A, ..., A, 2, ..., K
Event
Black:
2,
2, ...,
A
Event Union:
Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Color
Simple
Sample Space
Type
(S):
Ace
2, 2, 2,
..., A
Non-Ace
Total
Event
Ace  Black:
A,..., A, 2, ..., K
Total
Ace & Ace & Ace
Red
Black
Non & Non & NonRed
Black Ace
Red
Black
S
Red
Black
Simple Event Black:
2, ..., A
Event
Ace:
A,
A,
A,
A
Event Intersection:
Venn Diagram
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Sample
Space:
2, 2, 2,
..., A
Ace
Event Ace:
A, A, A, A
Black
S
Event Ace  Black:
A, A
Event
Black:
2,...,A
Event Intersection:
Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color &
Suit.
Color
Sample Space
Type
(S):
Ace
2, 2, 2,
..., A
Non-Ace
Event
Ace  Black:
A, A
Total
Total
Ace & Ace & Ace
Red
Black
Non & Non & NonRed
Black Ace
Red
Black
S
Red
Black
Simple
Event
Ace:
A, A,
A, A
Simple Event Black: 2, ..., A
Compound Event Probability
1. Numerical measure of likelihood that
compound event will occur
2. Can often use two–way table
• Two variables only
Event Probability Using
Two–Way Table
Event
Event
B1
B2
Total
A1
P(A 1  B1) P(A 1  B2) P(A 1)
A2
P(A 2  B1) P(A 2  B2) P(A 2)
Total
Joint Probability
P(B 1)
P(B 2)
1
Marginal (Simple) Probability
Two–Way Table Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Total
Ace
2/52
2/52
4/52
Non-Ace
24/52
24/52
48/52
Total
26/52
26/52
52/52
P(Red)
P(Ace  Red)
P(Ace)
Example from text book (p.145)
0.10+0.16+0.03+0.10+0.08+0.22+0.14=0.83
c. 0.16+0.03=0.19
P(A)=0.10+0.16+0.03=0.29
P(B)= 0.16+0.03+0.10+0.08+0.22+0.14=0.73
Thinking Challenge
What’s the Probability?
1. P(A) =
2. P(D) =
Event
C
D
4
2
3. P(C  B) =
Event
A
Total
6
4. P(A  D) =
B
1
3
4
5. P(B  D) =
Total
5
5
10
Solution*
The Probabilities Are:
1. P(A) = 6/10
2. P(D) = 5/10
Event
C
D
4
2
3. P(C  B) = 1/10
Event
A
Total
6
4. P(A  D) = 9/10
B
1
3
4
5. P(B  D) = 3/10 Total
5
5
10
3.3
Complementary Events
Complementary Events
Complement of Event A
• The event that A does not occur
• All events not in A
• Denote complement of A by AC
AC
A
S
Rule of Complements
The sum of the probabilities of complementary events
equals 1:
P(A) + P(AC) = 1
AC
A
S
Complement of Event
Example
Experiment: Draw 1 Card. Note Color.
Black
Sample
Space:
2, 2, 2,
..., A
Event Black:
2, 2, ..., A
S
Complement of Event Black,
BlackC: 2, 2, ..., A, A
Back to M&M example
a. S={Brown, yellow, red, blue, orange,green}
b. P={0.13, 0.14, 0.13, 0.24, 0.20, 0.16}
c. Let event A=selecting brown candy
P(A)=P(Brown)=0.13
d. Let event B=selecting red, green or yellow candy
P(B)= 0.13+0.16+0.14=0.43
e. Let event C= selecing a candy other than blue
P(C) = 0.13 +0.14+ 0.13+ 0.20+ 0.16=0.76
or P(C) = 1-0.24=0.76 P(Cc) = 0.24
3.4
The Additive Rule and
Mutually Exclusive Events
Mutually Exclusive Events
Mutually Exclusive Events
• Events do not occur
simultaneously
• A  B does not contain
any sample points

Mutually Exclusive
Events Example
Experiment: Draw 1 Card. Note Kind & Suit.
Sample
Space:
2, 2,
2, ..., A


Event Spade:
2, 3, 4, ..., A
S
Outcomes
in Event
Heart:
2, 3, 4
, ..., A
Events  and are Mutually Exclusive
Additive Rule
1. Used to get compound probabilities for union of
events
2.
P(A OR B) = P(A  B)
= P(A) + P(B) – P(A  B)
3. For mutually exclusive events:
P(A OR B) = P(A  B) = P(A) + P(B)
Additive Rule Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Ace
Red
Black
2
2
Total
4
Non-Ace
24
24
48
Total
26
26
52
P(Ace  Black) = P(Ace) + P(Black) – P(Ace  Black)
4
26
2 28
=
+
–
=
52
52 52 52
Thinking Challenge
Using the additive rule, what is the probability?
1. P(A  D) =
2. P(B  C) =
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
Solution*
Using the additive rule, the probabilities are:
1. P(A  D) = P(A) + P(D) – P(A  D)
6
5
2
9
=
+
–
=
10 10 10 10
2. P(B  C) = P(B) + P(C) – P(B  C)
4
5
1
8
=
+
–
=
10 10 10 10
Exercise from text book (p.152)
a.
b.
c.
d.
AB
Ac
BC
Ac Bc
Example from text book (p.153)
Example from text book (p.153): solution
Let events
• F= being a fully compensated worker
• R= being a partially compensated worker
• N=being a noncompasated volunteer
• L= leaving because of retirement
a. P(F)=127/244
b. P(FL)= 7/244
c. P(Fc)=1-(127/244)=117/244
d. P(FL)=P(F)+P(L)-P(FL)=(127/244)+(28/244)(7/244)=148/244