To Build or Not to Build (Part I)(review from last class)

Your business is located in a region that is somewhat prone to mud slides --- each rainy season there is a 1% chance of a mud slide occurring. You estimate that a mud slide would do $1,000,000 of damage.

You have the option of building a retaining wall that would help reduce the chance of a devastating mud slide. The wall costs $40,000 to build, and if a slide occurs, the wall will hold with a 95% probability.

Based on the EMC, should you build this wall?

To Build or Not to Build (Part II)

You also have the option of having a test done to determine whether or not a slide will occur in your specific location. The test costs $3,000 and has the following accuracies:

P(test positive | slide occurs) = 0.90P(test negative | slide doesn’t occur) = 0.85

If you choose the test, then, based on the result of the test, you will either build or not build.

Based on the EMC, what should your decision strategy be?

Risk in the Mud Slide Example

• EMC says to not build the wall• Is this the right decision?• There’s still a 1% chance of losing everything

($1,000,000 --- not to mention the business)

Given that you build the wall without testing, what is the probability that you’ll lose everything?

Given that you choose the test, what is the probability that you’ll lose everything?

Answers can be gotten from the tree…

Given that you build the wall without testing, what is the probability that you’ll lose everything?

With the given situation, there is one path (or sequence of events and decisions) that leads to losing everything:

Build w/o testing Slide Doesn’t Holdgiven 0.01 0.05

P( losing everything | build w/o testing ) = 0.01 * 0.05 = 0.0005

Given that you choose the test, what is the probability that you’ll lose everything?

With the given situation, there are two paths (or sequences of events and decisions) that lead to losing everything:

Test Pos Build Slide Doesn’t Holdgiven 0.1575 0.0571 0.05

Test Neg Don’t Build Slidegiven 0.8425 0.0012

P( first path ) = 0.1575 * 0.0571 * 0.05 = 0.00045

P( second path ) = 0.8425 * 0.0012 = 0.00101

P( losing everything | testing ) = 0.00045 + 0.00101 = 0.00146

Choice EMC “Risk” “Risk” fraction

Don’t Build 10.0000 0.01000 1 out of 100

Build w/o Testing 40.5000 0.00050 1 out of 2000

Test 10.7607 0.00146 1 out of 700

Summarizing our choices…

Now which looks best?

Perhaps testing is best because it balances both EMC and risk

Risk in Decision Analysis Problems

There are two ways to talk about risk in DA:

• Standard deviation (we’ll talk about this soon)

• “Negative outcome” analysis• Consider the probability of most negative outcome

• And balance it with EMC

Use negative outcome analysis on Homework #3

Sensitivity Analysis

We have analyzed the mud slide situation with the parameter that the wall costs $40,000. You may wonder: would my decision change if the wall cost $35,000 or $45,000 or …?

Sensitivity analysis is the investigation of how changes in the problem data

affect the problem outcome

“How sensitive is the outcome to changes in the data?”

Discrete Random Variables

discrete = consisting of unconnected, distinct parts

Think about a chance event with a finite number of outcomes…

A discrete random variable is an assignment of numbers, or values, to a collection of MECE events

MECE = mutually exclusive, collectively exhaustive

A

B

D

C

1.0

3.5

-2.0

3.5

• Usually, a capital letter, like X, is used to denote the random variable

• The assigned numbers mean something in the context of the situation (above is just an example)

New ideas for probability:

P( X = 1.0 ) = P(A)

P( X = 3.5 ) = P(B) + P(D)

P( X = -2.0 ) = P(C)

Examples of Discrete Random Variables

Heads and TailsX = 1 if Heads

X = 0 if Tails

P( X = 0 ) = P( X = 1 ) = 0.5

J&J Flea MarketSaturday

X = 1000 if No Rain P( X = 1000 ) = 0.3

X = 500 if Rain P( X = 500 ) = 0.7

J&J Flea MarketSunday

Y = 700 if No Rain P( Y = 700 ) = 0.7

Y = 350 if Rain P( Y = 350 ) = 0.3

Examples of Discrete Random Variables

StockX = price of stock

P( X = 50 ) = ???

Assembly Line

100 items produced per day

X = number of defective items

P( X = 20 ) = ???

P( X 5 ) = ???

P( X = 0 ) = ???

Random variables are often talked about in terms of the assigned values rather than in terms of the underlying events

Probability Mass Function(or “Probability Distribution”)

A probability mass function is the assignment of probabilities to the values of a discrete random variable

It is best understood as a bar graph

Investment A

Payoff Probability

-$2000 0.05

-$1000 0.35

$0 0.20

$1000 0.30

$2000 0.10

Expected Value of aDiscrete Random Variable

The expected value E[X] of a discrete random variable X is the average of all the possible values for X, weighted by the probabilities

Very similar to EMV and EMC…

Suppose X can take on the values a, b, and c. Then the expected value of X is

E[X] = a*P(X = a) + b*P(X = b) + c*P(X = c)

Examples of Expected Value

Heads and TailsP( X = 0 ) = P( X = 1 ) = 0.5

E[X] = 0*0.5 + 1*0.5 = 0.5

J&J Flea MarketSaturday

P( X = 1000 ) = 0.3 P( X = 500 ) = 0.7

E[X] = 0.3*1000 + 0.7*500 = 650

J&J Flea MarketSunday

P( Y = 700 ) = 0.7 P( Y = 350 ) = 0.3

E[Y] = 0.7*700 + 0.3*350 = 595

Same as EMV!

Examples of Expected Value

StockX = price of stock

P( X = 50 ) = ???

E[X] = ???

Assembly Line

100 items produced per day

X = number of defective items

P( X = 20 ) = ???

P( X 5 ) = ???

P( X = 0 ) = ???

E[X] = ???

Expected value is not always easy to calculate;we will learn some tricks for certain situations

Example of Expected Value

Investment A

Payoff Probability

-$2000 0.05

-$1000 0.35

$0 0.20

$1000 0.30

$2000 0.10

E[X] = -2 ( 0.05 ) + -1 ( 0.35 ) + 0 ( 0.20 ) + 1 ( 0.30 ) + 2 ( 0.10 )= 0.05

Standard Deviation of aDiscrete Random Variable

The standard deviation X is a measure of how far the values of a random variable X differ from (or deviate from) the expected value, E[X]

In a sense, you can think of E[X] as the “center” and X as a measure of how far the values of X are from the center

The bigger X , the bigger the deviation

If X takes values a, b, and c, then the standard deviation isX = square root of [ (a – E[X])2 P(X = a) +

(b – E[X])2 P(X = b) + (c – E[X])2 P(X = c) ]

Examples of Standard Deviation

Heads and TailsP( X = 0 ) = P( X = 1 ) = 0.5 E[X] = 0.5

X = sqrt [ (0 – 0.5)2 0.5 + (1 – 0.5)2 0.5 ]

= 0.5

J&J Flea MarketSaturday

P( X = 1000 ) = 0.3 P( X = 500 ) = 0.7

E[X] = 650 X = 229.13

J&J Flea MarketSunday

P( Y = 700 ) = 0.7 P( Y = 350 ) = 0.3

E[Y] = 595 X = 160.39

Examples of Standard Deviation

As with expected value, we will learn some tricks for calculating standard deviation for certain situations

Investment A

Payoff Probability

-$2000 0.05

-$1000 0.35

$0 0.20

$1000 0.30

$2000 0.10

Investment B

Payoff Probability

-$2000 0.15

-$1000 0.35

$0 0.00

$1000 0.30

$2000 0.20

Same EV, butB is riskier than A

The Binomial Distribution

1. One experiment is repeated n times2. In each experiment, only two outcomes are possible

--- either “success” or “failure”3. The experiments are independent, that is, knowing

the outcome of one gives you no information about the outcomes of the others

4. In each experiment, the probability of success is p (and so the probability of failure is 1-p)

5. The random variable X is “number of successes in n trials”

The binomial distribution is one of the most commonly encountered discrete random variable distributions. It has the following characteristics:

1. One experiment is repeated n times2. In each experiment, only two outcomes are possible --- either

“success” or “failure”3. The experiments are independent, that is, knowing the outcome of one

gives you no information about the outcome of the others4. In each experiment, the probability of success is p (and so the

probability of failure is 1-p)5. The random variable X is “number of successes in n trials”

The values of n and p are called the binomial parameters.

You have purchased 4 raffle tickets for a local 500-ticket raffle, and 10 prizes will be given away. What is the probability that you win at least 1 prize? What is the probability that you win exactly 2 prizes? What is the probability that you win 0 prizes?

n = 4, p = 10/500 = 0.02 P(X 1) P(X=2) P(X = 0)

How to Compute theBinomial Probabilities

Given n and p, we have a formula for P(X = k):

P(X = k) = nCk pk (1-p)n-k

nCk = “n choose k” = n! / ( k! (n-k)! )

7! = 7*6*5*4*3*2*13! = 3*2*1m! = m*(m-1)* … *3*2*10! = 1

Excel formula for P(X = k): BINOMDIST(k, n, p, 0)

Our Coin Flipping Experiment

Did it work?

Mean and Standard Deviationof the Binomial Distribution

E[X] = n * p

X = sqrt [ n * p * (1 – p) ]

Cumulative Binomial Probabilities

Useful for determining the probability of k or fewer successes:

P(X k)

P(X k) = P(X=0) + P(X=1) + P(X=2) + … + P(X=k)

Excel formula: BINOMDIST(k, n, p, 1)

Answers to Example

You have purchased 4 raffle tickets for a local 500-ticket raffle, and 10 prizes will be given away. What is the probability that you win at least 1 prize? What is the probability that you win exactly 2 prizes? What is the probability that you win 0 prizes?

Prob to win at least 1 prize = P(X 1) = 1 – P(X = 0) = 0.078

Prob to win no prizes = P(X = 0) = 0.922

Prob to win exactly 2 prizes = P(X = 2) = 0.002

Example for Binomial Distribution

A study by one automobile manufacturer indicated that one out of every four new cars required repairs under the company’s new-car warranty, with an average cost of $50 per repair.

a) For 100 new cars, what is the expected cost of repairs? What is the standard deviation?

b) Provide a reasonable estimate of the most it might cost the manufacturer for a specified group of 100 cars. (Hint: Ensure a 95% probability of not exceeding amount.)

Example for Binomial Distribution

Suppose you are in charge of hiring new undergraduate accounting majors for Coopers and Young (C&Y), an accounting firm headquartered in Chicago. This year your goal is to hire 20 graduates. On the average, about 40% of the people you make offers to will accept. Unfortunately, offers have to go out simultaneously this year, so you plan to make more than 20 offers.

a) How many offers should you make so that your expected number of hires is 20?

b) How many offers should you make if you want to have an 80% chance of hiring at least 20 people?

c) Find a 95% probability interval if you make 50 offers.

Review of Binomial Concepts

A single experiment, with probability of success p, is

repeated n independent times

X = number of successes

the discrete random variable

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

Probability distribution for X when n=10 and p=0.5

Cumulative distribution for X when n=10 and p=0.5

P(X = k)

P(X k)

P(X = k): BINOMDIST(k, n, p, 0)

P(X k): BINOMDIST(k, n, p, 1)

Excel Commands for Binomial Probabilities

X is a binomial random variable with binomial parameters n and p

Important Formulas forBinomial Random Variable E[X] = n * p

X = sqrt [ n * p * (1 – p) ]

Normal Distribution Example

Otis Elevator in Bloomington, Indiana, reported that the number of hours lost per week last year due to employees’ illnesses was approximately normally distributed, with a mean of 60 hours and a standard deviation of 15 hours. Determine, for a given week, the following probabilities:

a) The number of hours lost will exceed 85 hours.

b) The number of hours lost will be between 45 and 55 hours.