Transcript Document

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)
= 0.90
P(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
given
Slide
0.01
Doesn’t Hold
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
given
Build
0.0571
0.1575
Test
given
Slide
Neg
0.8425
Don’t Build
Doesn’t Hold
0.05
Slide
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
Summarizing our choices…
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
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
1.0
A
B
C
3.5
-2.0
D
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 = -2.0 ) = P(C)
P( X = 3.5 ) = P(B) + P(D)
Examples of Discrete Random Variables
X = 1 if Heads
Heads and Tails
X = 0 if Tails
P( X = 0 ) = P( X = 1 ) = 0.5
J&J Flea Market
Saturday
J&J Flea Market
Sunday
X = 1000 if No Rain P( X = 1000 ) = 0.3
X = 500 if Rain
P( X = 500 ) = 0.7
Y = 700 if No Rain
P( Y = 700 ) = 0.7
Y = 350 if Rain
P( Y = 350 ) = 0.3
Examples of Discrete Random Variables
X = price of stock
Stock
P( X = 50 ) = ???
100 items produced per day
X = number of defective items
Assembly Line
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 a
Discrete 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
P( X = 0 ) = P( X = 1 ) = 0.5
Heads and Tails
E[X] = 0*0.5 + 1*0.5 = 0.5
Same as EMV!
J&J Flea Market
Saturday
P( X = 1000 ) = 0.3
J&J Flea Market
Sunday
P( Y = 700 ) = 0.7
P( X = 500 ) = 0.7
E[X] = 0.3*1000 + 0.7*500 = 650
P( Y = 350 ) = 0.3
E[Y] = 0.7*700 + 0.3*350 = 595
Examples of Expected Value
X = price of stock
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 a
Discrete 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]
The bigger X , the bigger the deviation
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
If X takes values a, b, and c, then the standard deviation is
X = 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
P( X = 0 ) = P( X = 1 ) = 0.5 E[X] = 0.5
Heads and Tails
X = sqrt [ (0 – 0.5)2 0.5 + (1 – 0.5)2 0.5 ]
= 0.5
J&J Flea Market
Saturday
P( X = 1000 ) = 0.3
P( X = 500 ) = 0.7
E[X] = 650
X = 229.13
J&J Flea Market
Sunday
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
0.15
-$1000
0.35
$0
0.00
$1000
0.30
$2000
0.20
Payoff
Probability
-$2000
0.05
-$1000
0.35
$0
0.20
$1000
0.30
$2000
0.10
Investment B
Same EV, but
B is riskier than A
The Binomial Distribution
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 times
2. 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”
1.
2.
3.
4.
5.
One experiment is repeated n times
In each experiment, only two outcomes are possible --- either
“success” or “failure”
The experiments are independent, that is, knowing the outcome of one
gives you no information about the outcome of the others
In each experiment, the probability of success is p (and so the
probability of failure is 1-p)
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 the
Binomial 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*1
3! = 3*2*1
m! = m*(m-1)* … *3*2*1
0! = 1
Excel formula for P(X = k): BINOMDIST(k, n, p, 0)
Our Coin Flipping Experiment
Did it work?
Mean and Standard Deviation
of 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 no prizes = P(X = 0) = 0.922
Prob to win exactly 2 prizes = P(X = 2) = 0.002
Prob to win at least 1 prize = P(X  1) = 1 – P(X = 0) = 0.078
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
1
Probability
distribution
for X when
n=10 and
p=0.5
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
P(X = k)
0
0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
1
Cumulative
distribution
for X when
n=10 and
p=0.5
P(X  k)
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Excel Commands for Binomial Probabilities
X is a binomial random variable
with binomial parameters n and p
P(X = k): BINOMDIST(k, n, p, 0)
P(X  k): BINOMDIST(k, n, p, 1)
Important Formulas for
Binomial 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.