Introduction to Probability & Statistics Concepts of Probability
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Transcript of Introduction to Probability & Statistics Concepts of Probability
South Dakota
School of Mines & Technology
Introduction to Introduction to Probability & StatisticsProbability & Statistics
Industrial Engineering
Introduction to Probability & Statistics
Concepts of ProbabilityConcepts 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 dieS = {1, 2, 3, 4, 5, 6}
Body Temperature of a live humanS = [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} B = {2, 4, 6}
A B = {2}
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} B = {4, 5, 6}
A B = { } =
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) = 1/6p(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. 0 < P(A) < 12. P(S) = 13. P( ) = 04. 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
Alternatively we can find the probability of randomly selecting a technical person or a female by use of the addition rule.
= 18/30 + 14/30 - 10/30
= 22/30
Example (cont)
)()()()( FTPFPTPFTP -+=
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,TT,H T,T
P(H2|H1) = 1/2
Example Suppose we flip a fair coin twice. The
possible outcomes are:
H,H H,TT,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 followsTech Admin
Male
Female
8
10
8
4
Example
Let M = male, F = female, T = technical, and A = administrative. Compute the following:
P(M T) = ?
P(T|F) = ?
P(M|T) = ?
Tech Admin
Male
Female
8
10
8
4
South Dakota
School of Mines & Technology
Introduction to Introduction to Probability & StatisticsProbability & Statistics
Industrial Engineering
Introduction to Probability & Statistics
CountingCounting
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
2345
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
2345
345
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
2345
345
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,T,TH,H,T H,T,HH,T,H T,H,HT,H,H T,T,T
Combinations
Of 8 possible outcomes, 3 meet criteria
H,H,H H,T,TH,H,T H,T,HH,T,H T,H,HT,H,H T,T,T
Combinations
If we don’t care in which order these 3 occur
H,H,TH,T,HT,H,H
Then we can count by combination.
3 2
3
2 3 2
3 2 1
2 1 13C
!
!( )! ( )
Combinations
Combinations nCk = the number of ways to count k items out n total items order not important.
n = total number of itemsk = number of items pertaining to event A
k nCn
k n k
!
!( )!
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 4
10!
4 6!
10 9 8 7 6!
4 3 2 1 6!1 260C
!
,
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 2 12 3
8!
2 6!
12
3 96 160C C
!
!
! !,
Permutations
Permutations is somewhat like combinations except that order is important.
n kPn
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 4
10!
10 45 040P
( )!,
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 Introduction to Probability & StatisticsProbability & Statistics
Industrial Engineering
Introduction to Probability & Statistics
Random VariablesRandom 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
0 , headsX =
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
0 if TTTX = 1 if HTT, THT, TTH
2 if HHT, HTH, THH 3 if 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 1 2 3 4 5 6
p(x) 1/6 1/6
1/6 1/6
1/6 1/6
Cumulative
Let F(x) = Pr{X < x}
x 1 2 3 4 5 6
p(x) 1/6 1/6
1/6 1/6
1/6 1/6
F(x) 1/6 2/6
3/6 4/6
5/6 6/6
Complementary Cumulative
Let F(x) = 1 - F(x) = Pr{X > x}
x 1 2 3 4 5 6
p(x) 1/6 1/6
1/6 1/6
1/6 1/6
F(x) 1/6 2/6
3/6 4/6
5/6 6/6
F(x) 5/6 4/6
3/6 2/6
1/6 0/6
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,TH,H,T T,H,TH,T,H T,T,HT,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 trialsp = Pr(success) = const. from trial to trialn = number of trials
p(x) = b(x; n,p) =
n
x n xp px n x!
!( )!( )
1
Binomial Distribution
0.0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 4 5
x
P(x
)
0.0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 4 5
x
P(x
)
n=5, p=.3 n=8, p=.5
x
0.0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 4 5 6 7 8
P(x
)
n=4, p=.8
0.0
0.1
0.2
0.3
0.4
0.5
0 2 4
x
P(x
)
n=20, p=.5
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{ } Pr{ }X X 1 1 0
110
0 1005 95)0 10!
!( !)(. ) (.
= 1 - .9510
= .4013
Example
For p n Pr{X > 1}
.05 10 0.4013
.05 100 0.9941
.05 1,000 1.0000
.01 10 0.0956
.01 100 0.6340
.01 1,000 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
xa b c d
f(x)
A
1. f(x) > 0 , all x
2.
3. P(A) = Pr{a < x < b} =
4. Pr{X=a} =
f x dxa
d
( ) 1
f x dxb
c
( )f x dx
a
a
( ) 0
Continuous Univariate
Normal Uniform Exponential Weibull LogNormal
Beta T-distribution Chi-square F-distribution Maxwell Raleigh Triangular Generalized Gamma H-function
Normal Distribution
65%
95%
99.7%
f x eX
( )
FHG
IKJ1
2
1
2
2
Scale Parameter
x
> 1
= 1
Location Parameter
x
x
> 1
= 1
Std. Normal Transformation
Standard Normal
ZX
f(z)
N(0,1)f z e
z( )
1
2
1
22
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?
x
LSL USL
100
Example Cont.
x
LSL USL
100
Pr{in spec} = Pr{90 < x < 110}
Pr
90 100
5
110 100
5
x
= Pr(-2 < z < 2)
Example Cont.
x
LSL USL
100
Pr{in spec}= 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
Pr{poverty level} = Pr{X < 15,000}
= Pr{Z < -1.25}
= 0.5 - Pr{0 < Z < 1.25}
= 0.5 - 0.3944 = 0.1056
x
15,000 20,000
}000,4
000,20000,15Pr{
X
Other Continuous Distributions
Exponential Distribution
f x e x( ) Density
Cumulative
Mean 1/
Variance 1/2
F x e x( ) 1
, x > 0
0.0
0.5
1.0
0 0.5 1 1.5 2 2.5 3
Time to Fail
Den
sity
=1
Exponential Distribution
f x e x( ) Density
Cumulative
Mean 1/
Variance 1/2
F x e x( ) 1
, x > 0
=1
0.0
0.5
1.0
1.5
2.0
0 0.5 1 1.5 2 2.5 3
Time to Fail
Den
sity =2
0.0
0.5
1.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
x
f(x)
LogNormal
Density
Cumulative no closed form
Mean
Variance
f xx
ex
( )ln
1
2
1
2
2
, x > 0
e 2 2
e e2 2 2
1 ( ) = 0
=1
0.0
0.5
1.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
x
f(x)
LogNormal
Density
Cumulative no closed form
Mean
Variance
f xx
ex
( )ln
1
2
1
2
2
, x > 0
e 2 2
e e2 2 2
1 ( ) = 0
=2
-0.5
0.0
0.5
1.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
x
f(x)
LogNormal
Density
Cumulative no closed form
Mean
Variance
f xx
ex
( )ln
1
2
1
2
2
, x > 0
e 2 2
e e2 2 2
1 ( ) = 0
=0.5
Gamma
Density
Cumulative no closed form for integer
Mean
Variance 2
f x x e x( )( )
/
1 , x > 0
0.0
0.5
1.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
x
f(x) =1
0.0
0.5
1.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
x
f(x)
Gamma
Density
Cumulative no closed form for integer
Mean
Variance 2
f x x e x( )( )
/
1 , x > 0
=2
0.0
0.5
1.0
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
x
f(x)
Gamma
Density
Cumulative no closed form for integer
Mean
Variance 2
f x x e x( )( )
/
1 , x > 0
=3
0.0
0.5
1.0
1.5
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Weibull
Density
Cumulative
Mean Variance
f x x e x( ) ( / ) 2 1 2
, x > 0
F x e x( ) ( / ) 12
1
2 2
22 1 1
= 1
= 1
0.0
0.5
1.0
1.5
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Weibull
Density
Cumulative
Mean Variance
f x x e x( ) ( / ) 2 1 2
, x > 0
F x e x( ) ( / ) 12
1
2 2
22 1 1
= 1
= 2
0.0
0.5
1.0
1.5
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0
Weibull
Density
Cumulative
Mean Variance
f x x e x( ) ( / ) 2 1 2
, x > 0
F x e x( ) ( / ) 12
1
2 2
22 1 1
= 1
= 3
Uniform
Density
Cumulative
Mean (a + b)/2
Variance (b - a)2/12
f xb a
( )1
, a < x < b
F xx a
b a( )
f(x)
x
a b
End
Probability Review Session 1