Download Math/Stat 360-1 - WSU Department of Mathematics

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript
Math/Stat 360-1: Probability and Statistics,
Washington State University
Haijun Li
lih@math.wsu.edu
Department of Mathematics
Washington State University
Week 2
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
1 / 19
Outline
1
Section 2.2: Axioms, Interpretations, and Properties of
Probability
2
Section 2.3: Counting Techniques
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
2 / 19
Probability
Ω: Sample space.
Probability P(E): Likelihood of the random event E, E ⊆ Ω.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
3 / 19
Probability
Ω: Sample space.
Probability P(E): Likelihood of the random event E, E ⊆ Ω.
Basic Axioms of Probability Measures
1
2
3
0 ≤ P(E) ≤ 1 for all E ⊆ Ω.
P(Ω)
= 1.
mutually-exclusive-1.gif
(GIF Image, 411 × 275 pixels)
http://www.analyzemath.com
If events E1 , E2 , . . . , Ek , . . . , are
Pmutually exclusive,
P(E1 ∪ E2 ∪ · · · ∪ Ek ∪ · · · ) = ∞
i=1 P(Ei ).
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
3 / 19
Interpreting Probability
Probability P(E): Likelihood of the random event E (relative
frequency, or degree of belief, ...).
Figure: p ≈
Haijun Li
Pn
i=1 xi /n,
xi = 1 (head) xi = 0 (tail), n = number of trials
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
4 / 19
More Properties
P(A0 ) = 1 − P(A).
P(∅) = 0.
_diagram_example_0.png
(PNG Image,
484C,
× 243 pixels)
For any events
B and
P(B ∪ C) = P(B) + P(C) − P(B ∩ C).
Haijun Li
http://whyslopes.com/index
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
5 / 19
Example
Consider testing items coming off an assembly line one by one
until a defective item (labeled “F ”) is found. The sample space is
Ω = {F , SF , SSF , SSSF , SSSSF , . . . , S
. . S} F , . . . }.
| .{z
k ≥0
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
6 / 19
Example
Consider testing items coming off an assembly line one by one
until a defective item (labeled “F ”) is found. The sample space is
Ω = {F , SF , SSF , SSSF , SSSSF , . . . , S
. . S} F , . . . }.
| .{z
k ≥0
Probability Model
1
2
Let P(S) = 0.99, P(F ) = 0.01.
P(S
. . S} F ) = (0.99)k × 0.01.
| .{z
k ≥0
3
P(Ω) =
P∞
k =0
P(S
. . S} F ) =
| .{z
k ≥0
Haijun Li
∞
X
0.01
= 1.
1 − 0.99
k =0
|
{z
}
sum of a geometric series
(0.99)k 0.01 =
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
6 / 19
Example
In a city, 60% of all households get Internet service from the
local cable company, 80% get television service from that
company, and 50% get both services from that company. Let
A = {getting Internet service}, B = {getting TV service}.
1
What is the probability that a randomly selected household
gets at least one of these two services from the company?
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.8 − 0.5 = 0.9.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
7 / 19
Example
In a city, 60% of all households get Internet service from the
local cable company, 80% get television service from that
company, and 50% get both services from that company. Let
A = {getting Internet service}, B = {getting TV service}.
1
What is the probability that a randomly selected household
gets at least one of these two services from the company?
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.8 − 0.5 = 0.9.
2
What is the probability that a randomly selected household
gets exactly one of these services from the company?
P(exactly one) = P(A ∪ B) − P(A ∩ B) = 0.9 − 0.5 = 0.4.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
7 / 19
Venn Diagrams
Figure: P(A ∪ B), P(A ∩ B), P(exact one), P(A but not B)
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
8 / 19
Remarks
1
For any two events A ⊆ B, we have that P(A) ≤ P(B).
Because
P(B) = P(A ∪ (B ∩ A0 )) = P(A) + P(B ∩ A0 ) ≥ P(A).
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
9 / 19
Remarks
1
For any two events A ⊆ B, we have that P(A) ≤ P(B).
Because
P(B) = P(A ∪ (B ∩ A0 )) = P(A) + P(B ∩ A0 ) ≥ P(A).
2
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(B ∩ C) −
P(C ∩ A) + P(A ∩ B ∩ C).
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
9 / 19
Probability Model of Equally Likely Outcomes
Consider a sample space Ω = {ω1 , . . . , ωN }, consisting of N
sample points, N ≥ 1.
Assume that sample points ω1 , . . . , ωN are equally likely;
that is, P(ωi ) = N1 .
For any event A ⊆ Ω,
P(A) =
X
ωi ∈A
Haijun Li
P(ωi ) =
number of sample points in A
N(A)
=
.
N
N(Ω)
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
10 / 19
Probability Model of Equally Likely Outcomes
Consider a sample space Ω = {ω1 , . . . , ωN }, consisting of N
sample points, N ≥ 1.
Assume that sample points ω1 , . . . , ωN are equally likely;
that is, P(ωi ) = N1 .
For any event A ⊆ Ω,
P(A) =
X
ωi ∈A
P(ωi ) =
number of sample points in A
N(A)
=
.
N
N(Ω)
Example: Two fair dice are rolled separately. What is the
probability that the sum of the numbers is 4?
P(sum is 4) =
Haijun Li
3
1
=
.
36
12
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
10 / 19
The Product Rule for Ordered Pairs
Count all possible pairs (O, P)
The first element O can be selected in n1 ways.
The second element P can be selected in n2 ways.
The pair (O, P) can be selected in n1 n2 ways.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
11 / 19
Permutations and Combinations
Consider a set of n distinct objects, and 0 ≤ k ≤ n.
Definition
An ordered sequence of k objects is called a permutation.
Pk ,n = # of all permutations of size k from the n objects.
An unordered subset is called a combination.
Ck ,n = # of all combinations of size k from the n objects.
A more popular notation for the combination number is kn
(read “n choose k ”).
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
12 / 19
Figure: P2,3 = 6 = 2 × 3 = 2!C2,3 , C2,3 = 3
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
13 / 19
Pk ,n = k !Ck ,n , 0 ≤ k ≤ n
Consider a set of n distinct objects, and 0 ≤ k ≤ n.
Numbers of Permutations and Combinations
The n factorial n! = n(n − 1)(n − 2) · · · (2)(1). Note that
0! = 1.
Pk ,n = n(n − 1)(n − 2) · · · (n − k + 1)
|
{z
}
fill in k spots without repeating
n(n − 1)(n − 2) · · · (n − k + 1)(n − k )!
n!
=
(n − k )!
(n − k )!
P
Ck ,n = kk!,n = (n−kn!)!k ! = kn
n
, and n0 = nn = 1.
Note that kn = n−k
=
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
14 / 19
Example
Computer keyboard failures can be attributed to electrical
defects or mechanical defects. A repair facility currently has 25
failed keyboards, 6 of which have electrical defects and 19 of
which have mechanical defects.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
15 / 19
Example
Computer keyboard failures can be attributed to electrical
defects or mechanical defects. A repair facility currently has 25
failed keyboards, 6 of which have electrical defects and 19 of
which have mechanical defects.
How many ways are there to randomly select 5 of these
keyboards for a thorough inspection (with regard to order)?
total = P5,25 =
Haijun Li
25!
= 25×24×23×22×21 = 6, 375, 600.
(25 − 5)!
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
15 / 19
Example
Computer keyboard failures can be attributed to electrical
defects or mechanical defects. A repair facility currently has 25
failed keyboards, 6 of which have electrical defects and 19 of
which have mechanical defects.
How many ways are there to randomly select 5 of these
keyboards for a thorough inspection (with regard to order)?
total = P5,25 =
25!
= 25×24×23×22×21 = 6, 375, 600.
(25 − 5)!
How many ways are there to randomly select 5 of these
keyboards for a thorough inspection (without regard to
order)?
total = C5,25 =
Haijun Li
25!
25 × 24 × 23 × 22 × 21
=
= 53, 130.
20!5!
5!
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
15 / 19
Example (cont’d)
Computer keyboard failures can be attributed to electrical
defects or mechanical defects. A repair facility currently has 25
failed keyboards, 6 of which have electrical defects and 19 of
which have mechanical defects.
If a sample of 5 keyboards is randomly selected without
regard to order, what is the probability that exactly two have
an electrical defect?
Prob =
Haijun Li
N(A)
C2,6 C3,19
15 × 969
=
=
= 0.2736.
N
C5,25
53, 130
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
16 / 19
Example
A production facility employs 20 workers on the day shift, 15
workers on the swing shift, and 10 workers on the graveyard
shift. A quality control consultant is to select 6 of these workers
for in-depth interviews. Selections are made at random without
replacement.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
17 / 19
Example
A production facility employs 20 workers on the day shift, 15
workers on the swing shift, and 10 workers on the graveyard
shift. A quality control consultant is to select 6 of these workers
for in-depth interviews. Selections are made at random without
replacement.
How many selections result in all 6 workers coming from the
day shift?
20
20!
= 38, 760.
=
6
14!6!
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
17 / 19
Example
A production facility employs 20 workers on the day shift, 15
workers on the swing shift, and 10 workers on the graveyard
shift. A quality control consultant is to select 6 of these workers
for in-depth interviews. Selections are made at random without
replacement.
How many selections result in all 6 workers coming from the
day shift?
20
20!
= 38, 760.
=
6
14!6!
What is the probability that all 6 selected workers will be
from the day shift?
20
N(A)
6
= 0.0048.
= 45
Prob =
N(Ω)
6
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
17 / 19
Example (cont’d)
What is the probability that all 6 selected workers will be
from the same shift?
15
10
20
N(A)
6
6
6
Prob =
= 45 + 45 + 45 = 0.0054.
N(Ω)
6
6
6
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
18 / 19
Example (cont’d)
What is the probability that all 6 selected workers will be
from the same shift?
15
10
20
N(A)
6
6
6
Prob =
= 45 + 45 + 45 = 0.0054.
N(Ω)
6
6
6
What is the probability that at least two different shifts will
be represented among the selected workers?
Prob = 1 − Prob that all 6 will be from the same shift
= 1 − 0.0054 = 0.9946.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
18 / 19
Example (cont’d)
What is the probability that at least one of the shifts will be
unrepresented in the sample of workers?
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
19 / 19
Example (cont’d)
What is the probability that at least one of the shifts will be
unrepresented in the sample of workers?
A = {day shift will be unrepresented in the sample}
B = {swing shift will be unrepresented in the sample}
C = {graveyard shift will be unrepresented in the sample}
A ∩ B = {all 6 selected from the graveyard shift}
A ∩ C = {all 6 selected from the swing shift}
B ∩ C = {all 6 selected from the day shift}
A ∩ B ∩ C = ∅.
Haijun Li
Math/Stat 360-1: Probability and Statistics, Washington State University
Week 2
19 / 19
Example (cont’d)
What is the probability that at least one of the shifts will be
unrepresented in the sample of workers?
A = {day shift will be unrepresented in the sample}
B = {swing shift will be unrepresented in the sample}
C = {graveyard shift will be unrepresented in the sample}
A ∩ B = {all 6 selected from the graveyard shift}
A ∩ C = {all 6 selected from the swing shift}
B ∩ C = {all 6 selected from the day shift}
A ∩ B ∩ C = ∅.
P(A∪B∪C) = P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)
25
30
35
10
15
20
=
Haijun Li
6
45
6
+
6
45
6
+
6
45
6
−
6
45
6
−
6
45
6
−
6
45
6
Math/Stat 360-1: Probability and Statistics, Washington State University
= 0.2885.
Week 2
19 / 19
Document related concepts

Statistics wikipedia, lookup

History of statistics wikipedia, lookup

Probability wikipedia, lookup

Probability interpretations wikipedia, lookup

Ars Conjectandi wikipedia, lookup

Birthday problem wikipedia, lookup

Foundations of statistics wikipedia, lookup