Week 3 - Seminar

Report
Chapter 3: Probability
http://www.ocw.cn/OcwWeb/Sloan-School-of-Management/15-075AppliedStatisticsSpring2003/CourseHome/index.htm
Larson/Farber Ch. 3
Useful videos/websites:
 Video on terminology (outcome, event, experiment, sample space):
http://screencast.com/t/ODVlNzM2M2
 Video on 3 types of probability:
http://screencast.com/t/N2VkOTFlMz
 Website with good presentation on probability:
http://www.zweigmedia.com/RealWorld/tutorialsf15e/frames6_1.html
 Math Goodies:
http://www.mathgoodies.com/lessons/vol6/intro_probability.html
http://www.mathgoodies.com/lessons/vol6/independent_events.ht
ml
http://www.mathgoodies.com/lessons/vol6/conditional.html
http://www.mathgoodies.com/lessons/vol6/complement.html
Larson/Farber Ch. 3
Types of Probability
Classical (equally probable outcomes)
Empirical
Probability blood pressure will decrease
after medication
Intuition
Probability the line will be busy
Larson/Farber Ch. 3
Important Terms
Probability experiment:
An action through which counts, measurements or
responses are obtained
Sample space:
The set of all possible outcomes
Event:
A subset of the sample space.
Outcome:
The result of a single trial
Larson/Farber Ch. 3
Experiment: Flip a Coin and Roll a Die
Tree diagram:
H1 H2 H3 H4 H5 H6
T1 T2 T3 T4 T5 T6
The sample space has 12 outcomes:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Larson/Farber Ch. 3
Test yourself:
What’s the probability of the following events:
P(getting heads) =
b. P(rolling a 3) =
c. P(H3 or H4) =
d. P(rolling a Head AND getting a 3 or 4) =
a.
The sample space has 12 outcomes:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Larson/Farber Ch. 3
Fundamental Counting Principle
Fundamental Counting Principle
• If one event can occur in m ways and a second event can occur in
n ways, the number of ways the two events can occur in sequence
is m*n.
• Can be extended for any number of events occurring in
sequence.
Larson/Farber Ch. 3
Example
How many license plates can you make if a license number consists
of four alphabetic characters followed by two numbers?
__ __ __ __ __ __
Larson/Farber Ch. 3
Example
How many license plates can you make if a license number consists
of four unique alphabetic characters followed by two unique digits?
__ __ __ __ __ __
Larson/Farber Ch. 3
Definition: Complementary Events
Complement of event E [Denoted E ′ (E prime)]
The set of all outcomes in a sample space that are not included in
event E.
• P(E ′) + P(E) = 1
• P(E) = 1 – P(E ′)
• P(E ′) = 1 – P(E)
Larson/Farber Ch. 3
E
E′
Example
You survey a sample of
1000 employees at a
company and record the
age of each. Find the
probability of randomly
choosing an employee
who is not between 25
and 34 years old.
Employee ages
Frequency, f
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
Σf = 1000
Larson/Farber 4th ed
Larson/Farber Ch. 3
11
Solution – Complementary Events
 Event A = Age is between 25 and 34
 Event A’ = Age is NOT between 25 and 34
f
366
P(age 25 to 34)  
 0.366
n 1000
366
P(age is not 25 to 34)  1 
1000
634

 0.634
1000
Larson/Farber 4th ed
Larson/Farber Ch. 3
Employee ages
Frequency, f
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
Σf = 1000
12
Section 3.2
Conditional Probability
and the
Multiplication Rule
Larson/Farber Ch. 3
Conditional Probability
Conditional Probability
 The probability of an event occurring, given that another event has
already occurred
 Denoted P(B | A) (read “probability of B, given A”)
For Example:
 There are 6 frosted donuts and 6 plain donuts in a box. If you
select one at random, what’s the probability that you get a frosted
donut? [Simple probability. Answer: 1/2]
 Your friend takes a frosted donut, now what’s the probability that
you get a frosted donut? [Conditional probability. Answer: 5/11]
Larson/Farber Ch. 3
Example: Finding Conditional Probabilities
Two cards are selected in sequence from a standard deck. Find
the probability that the second card is a queen, given that the
first card is a king. (Assume that the king is not replaced.)
Event you are looking for: Second card is a queen.
Condition: You’ve already taken a king out of the deck.
15
Solution:
4
nd
st
Because
is a king
andisisa not
P( B | A) the
P(2first
cardcard
is a Queen
|1 card
Kingreplaced,
)
 0.078
51
the remaining deck has 51 cards, 4 of which are
queens.
Larson/Farber Ch. 3
Example: Finding Conditional Probabilities
The table shows the results of a study in which researchers
examined a child’s IQ and the presence of a specific gene in the
child. Find the probability that a child has a high IQ, given that the
child has the gene.
16
Gene
Present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
Larson/Farber 4th ed
Larson/Farber Ch. 3
Solution: Finding Conditional Probabilities
There are 72 children who have the gene. So, the sample space
consists of these 72 children.
Gene
Present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
Of these, 33 have a high IQ.
P( B | A)  P(high IQ | gene present ) 
17
Larson/Farber Ch. 3
33
 0.458
72
Independent and Dependent Events
Independent events
 The occurrence of one of the events does not affect the
probability of the occurrence of the other event
 P(B | A) = P(B) or P(A | B) = P(A)
 Events that are not independent are dependent
18
Larson/Farber Ch. 3
Larson/Farber 4th ed
Independent Events
Two events A and B are independent if the
probability of the occurrence of event B is not
affected by the occurrence
(or non-occurrence) of event A.
For example:
A = Being female
B = Having type O blood
A = 1st child is a boy
B = 2nd child is a boy
Larson/Farber Ch. 3
Dependent Events
Two events that are not independent are
dependent.
For example:
A = Living in Houston
B = Living in Texas
A = Selecting a red ball from (3 red, 3 blue)
B = Selecting a red ball, then a blue ball
Larson/Farber Ch. 3
Contingency Table
The results of responses when a sample of adults
in 3 cities was asked if they liked a new juice is:
Omaha
Yes
100
No
125
Undecided
75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Yes)
2. P(Seattle)
3. P(Miami)
4. P(No | Miami) = probability the answer is “no” given that
adult is a resident of Miami
Larson/Farber Ch. 3
Solutions
Yes
No
Undecided
Total
Omaha
100
125
75
300
1. P(Yes)
Seattle
150
130
170
450
Miami
150
95
5
250
= 400 / 1000 = 0.4
2. P(Seattle)
= 450 / 1000 = 0.45
3. P(Miami)
= 250 / 1000 = 0.25
4. P(No, given Miami)
= 95 / 250 = 0.38
Answers: 1) 0.4
Larson/Farber Ch. 3
Total
400
350
250
1000
2) 0.45
3) 0.25
4) 0.38
Imagine this
Two cars are selected from a production line of 12 where 5
are defective. Find the probability both cars are defective.
x
x
x
x
Larson/Farber Ch. 3
x
Multiplication Rule
To find the probability that two events, A and B will occur in
sequence, multiply the probability A occurs by the
conditional probability B occurs, given A has already
occurred.
A = first car is defective
B = second car is defective.
P(A and B) = P(A) x P(B|A)
P(A) = 5/12
P(B|A) = 4/11
P(A and B) = 5/12 x 4/11 = 5/33 = 0.1515
Larson/Farber Ch. 3
Multiplication Rule
Two dice are rolled. Find the probability both are 4’s.
A = first die is a 4 and B = second die is a 4.
P(A) = 1/6
P(B|A) = 1/6
P(A and B) = 1/6 x 1/6 = 1/36 = 0.028
When two events A and B are independent, then
P (A and B) = P(A) x P(B)
(because for independent events P(B) = P(B|A) )
Larson/Farber Ch. 3
.
Section 3.3
The Addition Rule
Larson/Farber Ch. 3
Compare “A and B” to “A or B”
The compound event “A and B” means that A
and B both occur in the same trial. Use the
multiplication rule to find P(A and B).
The compound event “A or B” means either A
can occur without B, B can occur without A or
both A and B can occur. Use the addition rule
to find P(A or B).
B
A
A and B
Larson/Farber Ch. 3
B
A
A or B
Mutually Exclusive Events
Two events, A and B, are mutually exclusive if
they cannot occur in the same trial.
A = A person is under 21 years old
B = A person is running for the U.S. Senate
A = A person was born in Philadelphia
B = A person was born in Houston
A
B
Mutually exclusive
P(A and B) = 0
When event A occurs it excludes event B in the same trial.
Larson/Farber Ch. 3
Non-Mutually Exclusive Events
If two events can occur in the same trial, they are
NOT mutually exclusive.
A = A person is under 25 years old
B = A person is a lawyer
A = A person was born in Philadelphia
B = A person watches “Jeopardy” on TV
A and B
Non-mutually exclusive
P(A and B) ≠ 0
Larson/Farber Ch. 3
A
B
The Addition Rule
The probability that one or the other of two events will
occur is:
P(A) + P(B) – P(A and B)
A card is drawn from a deck. Find the
probability it is a king or it is red.
Larson/Farber Ch. 3
The Addition Rule
The probability that one or the other of two events will
occur is:
P(A) + P(B) – P(A and B)
A card is drawn from a deck. Find the probability it is a king
or it is red.
A = the card is a king
B = the card is red.
P(A) = 4/52 and P(B) = 26/52
but P(A and B) = 2/52
P(A or B)
Larson/Farber Ch. 3
= 4/52 + 26/52 – 2/52
= 28/52 = 0.538
Example: Using the Addition Rule
A blood bank catalogs the types of blood given by donors during
the last five days. A donor is selected at random. Find the
probability the donor has type O or type A blood.
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
32
Larson/Farber Ch. 3
Larson/Farber 4th ed
Solution: Using the Addition Rule
The events are mutually exclusive (a donor cannot have type O
blood and type A blood)
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
P(type O or type A)  P(type O)  P(type A)
184 164


409 409
348

 0.851
409
33
Larson/Farber Ch. 3
Example: Using the Addition Rule
Find the probability the donor has type B or is Rh-negative.
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
Solution:
The events are not mutually exclusive (a donor can
have type B blood and be Rh-negative)
34
Larson/Farber Ch. 3
Larson/Farber 4th ed
Solution: Using the Addition Rule
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
P(type B or Rh  neg )
 P(type B)  P( Rh  neg )  P(type B and Rh  neg )
45
65
8
102




 0.249
409 409 409 409
35
Larson/Farber Ch. 3
Larson/Farber 4th ed
Section 3.4
Counting Principles
Larson/Farber Ch. 3
Permutations
Permutation
 An ordered arrangement of objects
 The number of different permutations of n distinct objects is n!
(n factorial)
 n! = n∙(n – 1)∙(n – 2)∙(n – 3)∙ ∙ ∙3∙2 ∙1
 0! = 1
 Examples:
 6! = 6∙5∙4∙3∙2∙1 = 720
 4! = 4∙3∙2∙1 = 24
37
Larson/Farber 4th ed
Larson/Farber Ch. 3
Permutations
Permutation of n objects taken r at a time
 The number of different permutations of n distinct objects taken
r at a time
■
n!
where r ≤ n
n Pr 
( n  r )!
38
Larson/Farber Ch. 3
Larson/Farber 4th ed
Example: Finding nPr
Find the number of ways of forming three-digit codes in which no
digit is repeated.
Solution:
You need select 3 digits from a group of 10
n = 10, r = 3
10!
10!

10 P3 
(10  3)!
7!
10  9  8  7  6  5  4  3  2  1

7  6  5  4  3  2 1
 720 ways
39
Larson/Farber Ch. 3
Larson/Farber 4th ed
Combinations
Combination of n objects taken r at a time
 A selection of r objects from a group of n objects without regard
to order
■
n
Cr
n!

( n  r )! r !
40
Larson/Farber Ch. 3
Larson/Farber 4th ed
Example: Combinations
A state’s department of transportation plans to develop a new
section of interstate highway and receives 16 bids for the project.
The state plans to hire four of the bidding companies. How many
different combinations of four companies can be selected from the
16 bidding companies?
Method:
• You need to select 4 companies from a group of 16
• n = 16, r = 4
• Order is not important
41
Larson/Farber Ch. 3
Larson/Farber 4th ed
Solution: Combinations
16!
16 C4 
(16  4)!4!
16!

12!4!
16 15 14 13 12!

12! 4  3  2 1
 1820 different combinations
42
Larson/Farber Ch. 3
Larson/Farber 4th ed
Larson/Farber Ch. 3
1. Two cars are selected from a production line
of 12 cars where 5 are defective. What is the
probability the 2nd car is defective, given the first
car was defective?
Larson/Farber Ch. 3
Contingency Table
The results of responses when a sample of adults in
3 cities was asked if they liked a new juice is:
Omaha
Yes
100
No
125
Undecided 75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
3. P(Miami or Yes)
2. P(Miami and Seattle) 4. P(Miami or Seattle)
Larson/Farber Ch. 3
3. One card is selected at random from a
standard deck, then replaced, and a second
card is drawn. Find the probability of
selecting two face cards.
A. 0.050
B. 0.053
C. 0.038
D. 0.462
Larson/Farber Ch. 3
4. The table shows the favorite pizza topping
for a sample of students. One of these
students is selected at random. Find the
probability the student is female or prefers
sausage.
Cheese Pepperoni Sausage Total
Male
8
5
2
15
A. 0.458
B. 0.583
Female
Total
2
10
4
9
3
5
9
24
C. 0.125
D. 0.556
Copyright © 2007 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley
Larson/Farber Ch. 3
Slide 3- 47
Answers
1. Given a defective car has been selected, the conditional sample space
has 4 defective out of 11. P(B|A) = 4/11
2. P(Miami and Yes) = 0.15
P(Miami and Seattle) = 0
P(Miami or Yes) = 250/1000 + 400/1000 – 150/1000 = 0.5
P(Miami or Seattle ) = 250/1000 + 450/1000 = 0.7
3. (B) 0.053
4. (A) .458
Larson/Farber Ch. 3

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