OPIM 5103-Lecture #3 Jose M. Cruz Assistant Professor.

OPIM 5103-Lecture #3

Jose M. Cruz

Assistant Professor

Introduction to Probability

• Counting Rules• Intro to Probability

– theoretical, empirical, subjective– probability rules!

• Probability Distributions– discrete: Uniform, Binomial, Poisson– continuous: Normal

A Counting Rule for Multiple-Step Experiments

If an experiment consists of a sequence of If an experiment consists of a sequence of kk steps steps in which there are in which there are nn11 possible results for the first step, possible results for the first step,

nn22 possible results for the second step, and so on, possible results for the second step, and so on,

then the total number of experimental outcomes isthen the total number of experimental outcomes is given by (given by (nn11)()(nn22) . . . () . . . (nnkk).).

A second useful counting rule enables us to count thenumber of experimental outcomes when n objects are tobe selected from a set of N objects.

Counting Rule for Combinations

CN

nN

n N nnN

!

!( )!C

N

nN

n N nnN

!

!( )!

Number of Number of CombinationsCombinations of of NN Objects Taken Objects Taken nn at a Time at a Time

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

Number of Number of PermutationsPermutations of of NN Objects Taken Objects Taken nn at a Time at a Time

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

P nN

nN

N nnN

!!

( )!P n

N

nN

N nnN

!!

( )!

CountingCounting Rule for Permutations Rule for Permutations

A third useful counting rule enables us to count theA third useful counting rule enables us to count thenumber of experimental outcomes when number of experimental outcomes when nn objects are to objects are tobe selected from a set of be selected from a set of NN objects, where the order of objects, where the order ofselection is important.selection is important.

Probability

Probability

• Probability is the numerical measure of the likelihood that an event will occur

• Value is between 0 and 1

• Sum of the probabilities of all mutually exclusive and collective exhaustive events is 1

Certain

Impossible

.5

1

0

(There are 2 ways to get one 6 and the other 4)e.g. P( ) = 2/36

Computing Probabilities

• The probability of an event E:

• Each of the outcomes in the sample space is equally likely to occur

number of event outcomes( )

total number of possible outcomes in the sample space

P E

X

T

Computing Joint Probability

• The probability of a joint event, A and B:

( and ) = ( )

number of outcomes from both A and B

total number of possible outcomes in sample space

P A B P A B

E.g. (Red Card and Ace)

2 Red Aces 1

52 Total Number of Cards 26

P

Computing Compound Probability

• Probability of a compound event, A or B:( or ) ( )

number of outcomes from either A or B or both

total number of outcomes in sample space

P A B P A B

E.g. (Red Card or Ace)

4 Aces + 26 Red Cards - 2 Red Aces

52 total number of cards28 7

52 13

P

Compound Probability (Addition Rule)

P(A or B ) = P(A) + P(B) - P(A and B)

For Mutually Exclusive Events: P(A or B) = P(A) + P(B)

P(A and B)P(A) P(B)

Computing Conditional Probability

• The probability of event A given that event B has occurred:

( and )( | )

( )

P A BP A B

P B

E.g.

(Red Card given that it is an Ace)

2 Red Aces 1

4 Aces 2

P

Conditional Probability

American Int’l Total

Men 0.25 0.15 0.40

Women 0.45 0.15 0.60

Total 0.70 0.30

Q: What is the probability that a randomly selected student is American, knowing that the student is female?

Conditional Probability and Statistical Independence

• Conditional probability:

• Multiplication rule:

( and )( | )

( )

P A BP A B

P B

( and ) ( | ) ( )

( | ) ( )

P A B P A B P B

P B A P A

Conditional Probability and Statistical Independence

• Events A and B are independent if

• Events A and B are independent when the probability of one event, A, is not affected by another event, B

(continued)

( | ) ( )

or ( | ) ( )

or ( and ) ( ) ( )

P A B P A

P B A P B

P A B P A P B

Binomial Probability Distribution

• ‘n’ identical trials– e.g.: 15 tosses of a coin; ten light bulbs taken from a

warehouse

• Two mutually exclusive outcomes on each trials– e.g.: Head or tail in each toss of a coin; defective or

not defective light bulb

• Trials are independent– The outcome of one trial does not affect the outcome

of the other

• Constant probability for each trial– e.g.: Probability of getting a tail is the same each time

we toss the coin

where:where: ff((xx) = the probability of ) = the probability of xx successes in successes in nn trials trials nn = the number of trials = the number of trials pp = the probability of success on any one trial = the probability of success on any one trial

( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x

Binomial DistributionBinomial Distribution

Binomial Probability FunctionBinomial Probability Function

Binomial Distribution

(1 )np p (1 )np p

EE((xx) = ) = = = npnp

Var(Var(xx) = ) = 22 = = npnp(1 (1 pp))

• Expected Value

• Variance

• Standard Deviation

Excel’s Binomial Function

=BINOMDIST(no. of successes, no. of trials, prob. of success, cumulative?)

Example=BINOMDIST(2,8,0.5, FALSE) (=0.11)

“Probability of tossing (exactly) two heads within 8 trials”

=BINOMDIST(2,8,0.5, TRUE) (=0.14)“Probability of tossing two heads or less within 8

trials”

Binomial Setting

Examples• Number of times newspaper arrives on time (i.e.,

before 7:30 AM) in a week/month• Number of times I roll “5” on a die in 20 rolls• Number of times I toss heads within 20 trials• Students pick random number between 1 and

10. Number of students who picked “7”• Number of people who will vote “Republican” in

a group of 20• Number of left-handed people in a group of 40

Service Center Staffing

0 0.36417 0.36417 Assumptions1 0.371602 0.735771 - 50 computers sold2 0.185801 0.921572 - Prob. customer calls for service = 0.023 0.06067 0.982242 - Want < 5% that there is no engineer4 0.014548 0.996795 0.002732 0.9995226 0.000418 0.999947 5.36E-05 0.9999948 5.88E-06 0.9999999 5.6E-07 1

10 4.69E-08 111 3.48E-09 112 2.31E-10 113 1.38E-11 114 7.42E-13 115 3.64E-14 116 1.62E-15 117 6.63E-17 118 2.48E-18 119 8.52E-20 120 2.7E-21 121 7.86E-23 122 2.11E-24 123 5.25E-26 124 1.21E-27 125 2.56E-29 126 5.02E-31 127 9.11E-33 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 5 10 15 20 25 30 35 40 45 50

Number of Service Calls

Pro

bab

ilit

y

Cumul. Prob.

Probability

Poisson Distribution

• Poisson Process:– Discrete events in an “interval”

• The probability of One Successin an interval is stable

• The probability of More thanOne Success in this interval is 0

– The probability of success isindependent from interval to interval

– e.g.: number of customers arriving in 15 minutes– e.g.: number of defects per case of light bulbs

P X x

x

x

( |

!

e-

Poisson DistributionPoisson Distribution

A property of the Poisson distribution is thatA property of the Poisson distribution is thatthe mean and variance are equal.the mean and variance are equal.

A property of the Poisson distribution is thatA property of the Poisson distribution is thatthe mean and variance are equal.the mean and variance are equal.

= = 22

Excel’s Poisson Function

=POISSON(no. of occurences, mean, cumulative?)

Example

=POISSON(5,2,FALSE) (=0.036)“Probability that (exactly) five customers arrive wihtin

an hour when the overall average is two”

=POISSON(5,2,TRUE) (=0.983)“Probability that five or less customers arrive wihtin an

hour when the overall average is two”

Poisson Setting

Examples• Number of accidents at an intersection in 6

months• Number of people entering a bank in a 30-

minute interval• Number of kids ringing the doorbell in 30

minutes for Halloween• Number of times a Microsoft machine crashes

within 24 hours• Number of sewing flaws per (100) garment(s)

Halloween

0 0.000335 0.000335 Assume: on average 4 kids /hour (=lambda)1 0.002684 0.0030192 0.010735 0.0137543 0.028626 0.042384 0.057252 0.0996325 0.091604 0.1912366 0.122138 0.3133747 0.139587 0.4529618 0.139587 0.5925479 0.124077 0.716624

10 0.099262 0.81588611 0.07219 0.88807612 0.048127 0.93620313 0.029616 0.96581914 0.016924 0.98274315 0.009026 0.99176916 0.004513 0.99628217 0.002124 0.99840618 0.000944 0.9993519 0.000397 0.99974720 0.000159 0.999906

A Poisson Distribution

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 2 3 4 5 6 7 8 9101112131415161718192021

Probability

Cum. Prob.

Continuous Probability Distributions

• Continuous random variable– Values from interval of numbers– Absence of gaps

• Continuous probability distribution– Distribution of continuous random variable

• Most important continuous probability distribution– The normal distribution

The Uniform Distribution

• “Rectangular shaped”• Every value between

a and b is equally likely

• The mean and median are in the middle

• Prob(X<=v) is the area on the left of v

Mean Median

X

f(X)

ba v

E(E(xx) = () = (aa + + bb)/2)/2E(E(xx) = () = (aa + + bb)/2)/2Var(Var(xx) = () = (bb - - aa))22/12/12Var(Var(xx) = () = (bb - - aa))22/12/12

The Normal Distribution

• “Bell shaped”• Symmetrical• Mean, median and

mode are equal• Interquartile range

equals 1.33 • 68-95-99 % rule• Random variable

has infinite range

Mean Median

X

f(X)

Generating Random VariablesUsing Excel

• Excel can be used to generate Discrete and Continuous Random Variables

• Complex Probabilistic Models can be constructed and simulation can give insight and suggest managerial decisions

The Mathematical Model

21

2

2

1

2

: density of random variable

3.14159; 2.71828

: population mean

: population standard deviation

: value of random variable

X

f X e

f X X

e

X X

Many Normal Distributions

By varying the parameters and , we obtain different normal distributions

There are an infinite number of normal distributions

Finding Probabilities

Probability is the area under the curve!

c dX

f(X)

?P c X d

Which Table to Use?

An infinite number of normal distributions means an infinite number of tables to look

up!

Solution: The Cumulative Standardized Normal Distribution

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.5478.02

0.1 .5478

Cumulative Standardized Normal Distribution Table (Portion)

Probabilities

Shaded Area Exaggerated

Only One Table is Needed

0 1Z Z

Z = 0.12

0

Standardizing Example

6.2 50.12

10

XZ

Normal Distribution

Standardized Normal

Distribution


10 1Z

5 6.2 X Z0Z

0.12

Example:

Normal Distribution

Standardized Normal

Distribution


10 1Z

5 7.1 X Z0Z

0.21

2.9 5 7.1 5.21 .21

10 10

X XZ Z

2.9 0.21

.0832

2.9 7.1 .1664P X

.0832

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.5832.02

0.1 .5478



0 1Z Z

Z = 0.21

Example:

2.9 7.1 .1664P X (continued)

0

Z .00 .01

-03 .3821 .3783 .3745

.4207 .4168

-0.1.4602 .4562 .4522

0.0 .5000 .4960 .4920

.4168.02

-02 .4129



0 1Z Z

Z = -0.21

Example:

2.9 7.1 .1664P X (continued)

0

Example:

8 .3821P X

Normal Distribution

Standardized Normal

Distribution


10 1Z

5 8 X Z0Z

0.30

8 5.30

10

XZ

.3821

Example:

8 .3821P X (continued)

Z .00 .01

0.0 .5000 .5040 .5080

.5398 .5438

0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255

.6179.02

0.1 .5478



0 1Z Z

Z = 0.30

0

.6217

Finding Z Values for Known Probabilities

Z .00 0.2

0.0 .5000 .5040 .5080

0.1 .5398 .5438 .5478

0.2 .5793 .5832 .5871

.6179 .6255

.01

0.3

Cumulative Standardized Normal Distribution Table

(Portion)

What is Z Given Probability = 0.1217 ?


.6217

0 1Z Z

.31Z 0

Recovering X Values for Known Probabilities

5 .30 10 8X Z

Normal Distribution

Standardized Normal

Distribution10 1Z

5 ? X Z0Z 0.30

.3821.1179

Finding Probabilities for X Values Using Excel

Excel function:

=NORMDIST(x,mean,standard_deviation,TRUE)

=NORMSDIST(z,TRUE)

Example

Prob.(weight <= 165 lbs) when mean=180, std_dev=20:

=NORMDIST(165,180,20,true)Answer: 0.2267

Prob.(weight >= 185 lbs) ?

Prob.(weight >= 165 and weight <= 185 lbs) ?

Finding X Values for Known ProbabilitiesUsing Excel

Excel function:

=NORMINV(probabiltiy,mean,standard_deviation)

=NORMSINV(probability)

Example

Prob.(weight <= X)= 0.2 (mean=180, std_dev=20)

=NORMINV(0.2,180,20)Answer: X=163

Prob.(weight >= X)=0.4 X?Answer: X=185

Assessing Normality

• Not all continuous random variables are normally distributed

• It is important to evaluate how well the data set seems to be adequately approximated by a normal distribution

Assessing Normality

• Construct charts– For large data sets, does the histogram appear bell-

shaped?

• Compute descriptive summary measures– Do the mean, median and mode have similar values?– Is the interquartile range approximately 1.33 ?– Does the data obey the 68-95-99 percent rule?– Is the range approximately 6 ?

(continued)

Assessing Normality

• Observe the distribution of the data set– Do approximately 2/3 of the observations lie

between mean 1 standard deviation?– Do approximately 4/5 of the observations lie

between mean 1.28 standard deviations?– Do approximately 19/20 of the observations lie

between mean 2 standard deviations?

(continued)

OPIM 5103-Lecture #3 Jose M. Cruz Assistant Professor.

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Transcript of OPIM 5103-Lecture #3 Jose M. Cruz Assistant Professor.