Transcript Prob. Review I

South Dakota
School of Mines & Technology
Introduction to
Probability & Statistics
Industrial Engineering
Introduction to
Probability & Statistics
Concepts of Probability
Probability Concepts
S = Sample Space : the set of all possible unique
outcomes of a repeatable experiment.
Ex:
flip of a coin
S = {H,T}
No. dots on top face of a die
S = {1, 2, 3, 4, 5, 6}
Body Temperature of a live human
S = [88,108]
Probability Concepts
Event: a subset of outcomes from a sample
space.
Simple Event: one outcome; e.g. get a 3 on one
throw of a die
A = {3}
Composite Event: get 3 or more on throw of a
die
A = {3, 4, 5, 6}
Rules of Events
Union: event consisting of all outcomes present
in one or more of events making up
union.
Ex:
A = {1, 2} B = {2, 4, 6}
A  B = {1, 2, 4, 6}
Rules of Events
Intersection: event consisting of all outcomes
present in each contributing event.
Ex:
A = {1, 2}
A  B = {2}
B = {2, 4, 6}
Rules of Events
Complement: consists of the outcomes in the
sample space which are not in stipulated
event
Ex:
A = {1, 2}
S = {1, 2, 3, 4, 5, 6}
A = {3, 4, 5, 6}
Rules of Events
Mutually Exclusive: two events are mutually
exclusive if their intersection is null
Ex:
A = {1, 2, 3}
AB={ }=
B = {4, 5, 6}
Probability Defined
 Equally
Likely Events
If m out of the n equally likely outcomes in an
experiment pertain to event A, then
p(A) = m/n
Probability Defined
 Equally
Likely Events
If m out of the n equally likely outcomes in an
experiment pertain to event A, then
p(A) = m/n
Ex: Die example has 6 equally likely
outcomes:
p(2) = 1/6
p(even) = 3/6
Probability Defined
 Suppose
we have a workforce which is
comprised of 6 technical people and 4 in
administrative support.
Probability Defined
 Suppose
we have a workforce which is
comprised of 6 technical people and 4 in
administrative support.
P(technical) = 6/10
P(admin) = 4/10
Rules of Probability
Let A = an event defined on the event space S
1.
2.
3.
4.
0 < P(A) < 1
P(S) = 1
P( ) = 0
P(A) + P( A ) = 1
Addition Rule
P(A B) = P(A) + P(B) - P(A  B)
A
B
Addition Rule
P(A B) = P(A) + P(B) - P(A  B)
A
B
Example
 Suppose
we have technical and
administrative support people some of whom
are male and some of whom are female.
Example (cont)
 If
we select a worker at random, compute the
following probabilities:
P(technical) = 18/30
Example (cont)
 If
we select a worker at random, compute the
following probabilities:
P(female) = 14/30
Example (cont)
 If
we select a worker at random, compute the
following probabilities:
P(technical or female) = 22/30
Example (cont)
 If
we select a worker at random, compute the
following probabilities:
P(technical and female) = 10/30
Example (cont)
 Alternatively
we can find the probability of
randomly selecting a technical person or a
female by use of the addition rule.
P (T  F ) = P (T ) + P ( F ) - P (T  F )
= 18/30 + 14/30 - 10/30
= 22/30
Operational Rules
Mutually Exclusive Events:
P(A B) = P(A) + P(B)
A
B
Conditional Probability
Now suppose we know that event A has
occurred. What is the probability of B given A?
A
AB
P(B|A) = P(A  B)/P(A)
Example
 Returning
to our workers, suppose we know
we have a technical person.
Example
 Returning
to our workers, suppose we know
we have a technical person. Then,
P(Female | Technical) = 10/18
Example
 Alternatively,
P(F | T) = P(F  T) / P(T)
= (10/30) / (18/30) = 10/18
Independent Events
 Two
events are independent if
P(A|B) = P(A)
or
P(B|A) = P(B)
In words, the probability of A is in no way
affected by the outcome of B or vice versa.
Example
 Suppose
we flip a fair coin. The possible
outcomes are
H
T
The probability of getting a head is then
P(H) = 1/2
Example
 If
the first coin is a head, what is the
probability of getting a head on the second
toss?
H,H H,T
T,H T,T
P(H2|H1) = 1/2
Example
 Suppose
we flip a fair coin twice. The
possible outcomes are:
H,H H,T
T,H T,T
P(2 heads) = P(H,H) = 1/4
Example
 Alternatively
P(2 heads) = P(H1  H2)
= P(H1)P(H2|H1)
= P(H1)P(H2)
= 1/2 x 1/2
= 1/4
Example
 Suppose
we have a workforce consisting of
male technical people, female technical
people, male administrative support, and
female administrative support. Suppose the
make up is as follows
Tech
Admin
Male
8
8
Female
10
4
Example
Let M = male, F = female, T = technical, and
A = administrative. Compute the following:
Tech
Male
Female
Admin
8
8
10
4
P(M  T) = ?
P(T|F) = ?
P(M|T) = ?
South Dakota
School of Mines & Technology
Introduction to
Probability & Statistics
Industrial Engineering
Introduction to
Probability & Statistics
Counting
Fundamental Rule
 If
an action can be performed in m ways and
another action can be performed in n ways,
then both actions can be performed in m•n
ways.
Fundamental Rule
 Ex:
A lottery game selects 3 numbers
between 1 and 5 where numbers can not be
selected more than once. If the game is truly
random and order is not important, how
many possible combinations of lottery
numbers are there?
Fundamental Rule

Ex: A lottery game selects 3 numbers between 1 and 5 where
numbers can not be selected more than once. If the game is
truly random and order is not important, how many possible
combinations of lottery numbers are there?
1
2
3
4
5
Fundamental Rule

Ex: A lottery game selects 3 numbers between 1 and 5 where
numbers can not be selected more than once. If the game is
truly random and order is not important, how many possible
combinations of lottery numbers are there?
1
2
3
4
5
2
3
4
5
Fundamental Rule

Ex: A lottery game selects 3 numbers between 1 and 5 where
numbers can not be selected more than once. If the game is
truly random and order is not important, how many possible
combinations of lottery numbers are there?
1
2
3
4
5
2
3
4
5
3
4
5
Fundamental Rule

Ex: A lottery game selects 3 numbers between 1 and 5 where
numbers can not be selected more than once. If the game is
truly random and order is not important, how many possible
combinations of lottery numbers are there?
1
2
3
4
5
2
3
4
5
3
4
5
LN = 5•4•3
= 60
Combinations
 Suppose
we flip a coin 3 times, how many
ways are there to get 2 heads?
Combinations
 Suppose
we flip a coin 3 times, how many
ways are there to get 2 heads?
Soln:
List all possibilities:
H,H,H
H,H,T
H,T,H
T,H,H
H,T,T
H,T,H
T,H,H
T,T,T
Combinations
Of 8 possible outcomes, 3 meet criteria
H,H,H
H,H,T
H,T,H
T,H,H
H,T,T
H,T,H
T,H,H
T,T,T
Combinations
If we don’t care in which order these 3 occur
H,H,T
H,T,H
T,H,H
Then we can count by combination.
3!
3  2 1

3
3 C2 
2 !(3  2)! 2  1 (1)
Combinations
 Combinations nCk
= the number of ways to
count k items out n total items order not
important.
n!
k Cn 
k !(n  k )!
n = total number of items
k = number of items pertaining to event A
Example
 How
many ways can we select a 4 person
committee from 10 students available?
Example
 How
many ways can we select a 4 person
committee from 10 students available?
No. Possible Committees =
10! 10  9  8  7  6!

 1,260
10 C4 
4! 6! 4  3  2  1 6!
Example
 We
have 20 students, 8 of whom are female
and 12 of whom are male. How many
committees of 5 students can be formed if we
require 2 female and 3 male?
Example
 We
have 20 students, 8 of whom are female
and 12 of whom are male. How many
committees of 5 students can be formed if we
require 2 female and 3 male?
Soln:
Compute how many 2 member female
committees we can have and how many 3
member male committees. Each female
committee can be combined with each male
committee.
Example
8!
12 !

 6,160
8 C2 12 C3 
2 ! 6! 3! 9 !
Permutations
 Permutations
is somewhat like combinations
except that order is important.
n!
n Pk 
(n  k )!
Example
 How
many ways can a four member
committee be formed from 10 students if the
first is President, second selected is Vice
President, 3rd is secretary and 4th is
treasurer?
Example

How many ways can a four member committee be
formed from 10 students if the first is President,
second selected is Vice President, 3rd is secretary and
4th is treasurer?
10!
 5,040
10 P4 
(10  4)!
Example

How many ways can a four member committee be
formed from 10 students if the first is President,
second selected is Vice President, 3rd is secretary and
4th is treasurer?
10P4
•
•
= 10*9*8*7 = 5,040
South Dakota
School of Mines & Technology
Introduction to
Probability & Statistics
Industrial Engineering
Introduction to
Probability & Statistics
Random Variables
Random Variables
A Random Variable is a function that
associates a real number with each element in a
sample space.
Ex:
Toss of a die
X = # dots on top face of die
= 1, 2, 3, 4, 5, 6
Random Variables
A Random Variable is a function that
associates a real number with each element in a
sample space.
Ex:
Flip of a coin
X=

0 , heads
1 , tails
Random Variables
A Random Variable is a function that
associates a real number with each element in a
sample space.
Ex:
Flip 3 coins

X=
0
1
2
3
if
if
if
if
TTT
HTT, THT, TTH
HHT, HTH, THH
HHH
Random Variables
A Random Variable is a function that
associates a real number with each element in a
sample space.
Ex:
X = lifetime of a light bulb
X = [0, )
Distributions
Let X = number of dots on top face of a die
when thrown
p(x) = Prob{X=x}
x
p(x)
1
2
3
4
5
6
1/ 1/ 1/ 1/ 1/ 1/
6
6
6
6
6
6
Cumulative
Let F(x) = Pr{X < x}
x
1
2
3
4
5
6
p(x)
1/
F(x)
1/ 2/ 3/ 4/ 5/ 6/
6
6
6
6
6
6
1/ 1/ 1/ 1/ 1/
6
6
6
6
6
6
Complementary Cumulative
Let F(x) = 1 - F(x) = Pr{X > x}
x
1
2
3
4
5
6
p(x)
1/ 1/ 1/ 1/ 1/ 1/
6
6
6
6
6
6
F(x)
1/ 2/ 3/ 4/ 5/ 6/
6
6
6
6
6
6
F(x)
5/ 4/ 3/ 2/ 1/ 0/
6
6
6
6
6
6
Discrete Univariate
 Binomial
 Discrete
Uniform (Die)
 Hypergeometric
 Poisson
 Bernoulli
 Geometric
 Negative
Binomial
Binomial
 What
is the probability of getting 2 heads out
of 3 flips of a coin?
Binomial
 What
is the probability of getting 2 heads out
of 3 flips of a coin?
Soln:
H,H,H
H,T,T
H,H,T
T,H,T
H,T,H
T,T,H
T,H,H
T,T,T
Binomial
P{2 heads in 3 flips} = P{H,H,T} + P{H,T,H}
+ P{T,H,H}
= 3•P{H}P{H}P{T}
= 3C2•P{H}2•P{T}3-2
= 3C2•p2•(1-p)3-2
Distributions
Binomial:
X = number of successes in n bernoulli trials
p = Pr(success) = const. from trial to trial
n = number of trials
n!

 x
n x
p
(
1
p
)


p(x) = b(x; n,p) = 
 x !(n  x)!
Binomial Distribution
n=8, p=.5
0.5
0.5
0.4
0.4
0.3
0.3
P(x)
P(x)
n=5, p=.3
0.2
0.1
0.2
0.1
0.0
0.0
0
1
2
3
4
5
0
1
x
4
0.4
0.3
P(x)
0.3
0.2
0.1
0.2
0.1
0.0
3
4
5
x
6
7
8
4
2
2
1
0
0.0
0
5
n=20, p=.5
0.5
0.4
P(x)
3
x
n=4, p=.8
0.5
2
x
Example

Suppose we manufacture circuit boards with 95%
reliability. If approximately 5 circuit boards in 100
are defective, what is the probability that a lot of 10
circuit boards has one or more defects?
Example (soln.)
Pr{X  1}  1  Pr{X  0}
 10 ! 
0
10
1 
 (.05) (.95)
 0 !(10 !)
= 1 - .9510
= .4013
Example
For
p
.05
.05
.05
.01
.01
.01
n
Pr{X > 1}
10
100
1,000
10
100
1,000
0.4013
0.9941
1.0000
0.0956
0.6340
1.0000
99% Defect Free Rate
500 incorrect surgical procedures every week
 20,000 prescriptions filled incorrectly each
year
 12 babies given to the wrong parents each
day
 16,000 pieces of mail lost each hour
 2 million documents lost by IRS each year
 22,000 checks deducted from wrong accounts
during next hour

(Ref: Quality, March 91)
Continuous Distribution
f(x)
A
x
a
1. f(x) > 0
b
c
,
d
all x
d
2.  f ( x)dx  1
a
c
3. P(A) = Pr{a < x < b} =  f ( x)dx
a
4. Pr{X=a} =  f ( x)dx  0
a
b
Continuous Univariate





Normal
Uniform
Exponential
Weibull
LogNormal









Beta
T-distribution
Chi-square
F-distribution
Maxwell
Raleigh
Triangular
Generalized Gamma
H-function
Normal Distribution
1
f ( x) 
2
F
IJ
G
e H K
1 X 

2 
  
2

65%
95%
99.7%



Scale Parameter
>1
x
  




=1



Location Parameter
x
>1

x

=1
Std. Normal Transformation
f(z)
Standard Normal
X 
Z


1
f ( z) 
e
2
1
 z2
2

N(0,1)

Example
 Suppose
a resistor has specifications of 100 +
10 ohms. R = actual resistance of a resistor
and R
N(100,5). What is the probability a
resistor taken at random is out of spec?

LSL
USL
x

100

Example Cont.
LSL
USL
x

Pr{in spec}
100

= Pr{90 < x < 110}
90  100 x   110  100


 Pr 




5
5

= Pr(-2 < z < 2)
Example Cont.
LSL
USL
x

Pr{in spec}
100

= Pr(-2 < z < 2)
= [F(2) - F(-2)]
= (.9773 - .0228) = .9545
Pr{out of spec} = 1 - Pr{in spec}
= 1 - .9545
= 0.0455
Example
 Assume
that the per capita income in South
Dakota is normally distributed with a mean
of $20,000 and a standard deviation of $4,000.
If the poverty level is considered to be $15,000
per year, compute the percentage of South
Dakotans who would be considered to be at
or below the poverty level.
Example
x
15,000
20,000
Pr{poverty level} = Pr{X < 15,000}
X   15,000  20,000
 Pr{

}

4,000
= Pr{Z < -1.25}
= 0.5 - Pr{0 < Z < 1.25}
= 0.5 - 0.3944 = 0.1056
Other Continuous
Distributions
Exponential Distribution
Density
f ( x )  e
Cumulative
F ( x)  1  e
Mean
1/
Variance
1/2
 x
,x>0
 x
1.0
Density
=1
0.5
0.0
0
0.5
1
1.5
Time to Fail
2
2.5
3
Exponential Distribution
Density
f ( x )  e
Cumulative
F ( x)  1  e
Mean
1/
Variance
1/2
 x
,x>0
 x
2.0
Density
1.5
=1
=2
1.0
0.5
0.0
0
0.5
1
1.5
Time to Fail
2
2.5
3
LogNormal
1
Density
f ( x) 
Cumulative
no closed form
Variance
2
,x>0
  2 2
e
e 2    ( e  1)
2
=0
2
1.0
f(x)
Mean
x 2
e
1  ln x   
 

2  
=1
0.5
0.0
0.0
1.0
2.0
3.0
4.0
x
5.0
6.0
7.0
LogNormal
1
Density
f ( x) 
Cumulative
no closed form
Variance
2
,x>0
  2 2
e
e 2    ( e  1)
2
=0
2
1.0
f(x)
Mean
x 2
e
1  ln x   
 

2  
=2
0.5
0.0
0.0
1.0
2.0
3.0
4.0
x
5.0
6.0
7.0
LogNormal
1
Density
f ( x) 
Cumulative
no closed form
Variance
2
,x>0
  2 2
e
e 2    ( e  1)
2
=0
2
1.0
=0.5
0.5
f(x)
Mean
x 2
e
1  ln x   
 

2  
0.0
0.0
1.0
2.0
3.0
4.0
-0.5
x
5.0
6.0
7.0
Gamma
Density
     1  x/
f ( x) 
x e
 ( )
,x>0
Cumulative
no closed form for   integer
Mean

Variance
2
f(x)
1.0
=1
0.5
0.0
0.0
1.0
2.0
3.0
4.0
x
5.0
6.0
7.0
Gamma
Density
     1  x/
f ( x) 
x e
 ( )
,x>0
no closed form for   integer
Mean

Variance
2
f(x)
Cumulative
1.0
=2
0.5
0.0
0.0
1.0
2.0
3.0
4.0
x
5.0
6.0
7.0
Gamma
Density
     1  x/
f ( x) 
x e
 ( )
,x>0
Cumulative
no closed form for   integer
Mean

Variance
2
1.0
f(x)
=3
0.5
0.0
0.0
1.0
2.0
3.0
4.0
x
5.0
6.0
7.0
Weibull
Density
f ( x)   x
2
Cumulative F ( x)  1  e
e
 ( x/ ) 2
,x>0
 ( x/ ) 2
2
Variance

  1
 
  
Mean
 1
  2 1  1 
 2          


2
1.5
= 1
1.0
=1
0.5
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Weibull
f ( x)   x
Density
2
Cumulative F ( x)  1  e
e
 ( x/ ) 2
,x>0
 ( x/ ) 2
2
Variance

  1
 
  
Mean
 1
  2 1  1 
 2          


2
1.5
= 1
1.0
=2
0.5
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Weibull
f ( x)   x
Density
2
Cumulative F ( x)  1  e
e
 ( x/ ) 2
,x>0
 ( x/ ) 2
2
Variance

  1
 
  
Mean
 1
  2 1  1 
 2          


2
1.5
= 1
=3
1.0
0.5
0.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Uniform
Density
Cumulative
1
f ( x) 
b a
x a
F ( x) 
b a
Mean
(a + b)/2
Variance
(b - a)2/12
, a<x<b
f(x)
x
a
b
End
Probability Review
Session 1