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

Math/Stat 360-1: Probability and Statistics,
Washington State University
Haijun Li
[email protected]
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