Docstoc

Probability

Document Sample
Probability Powered By Docstoc
					Lecture 2
Probability
  – some examples
  – sources of bias
  – types of probability
Problems
Expected Values
Conditional Probability
Important probability formulas
An exercise first
Answer the questions on the sheet of paper
 that is handed to you (should take less than
 1 minute).
You’re not being tested on your general
 knowledge -- I just want to get your
 estimate.
Excel functions last week
Calculate:
  –   mean using =AVERAGE()
  –   median using =MEDIAN()
  –   mode using =MODE()
  –   standard deviation using =STDEV()
  –   variance using =VAR()
  –   Get all these at once using >Tools>Data
      Analysis> Descriptive statistics
Some thought exercises
  about Probability
Birthday Problem

What is the chance of two people in this
 class having the same birthday?
Causes of Death

Which is more likely: being killed by
 aeroplane parts falling from the sky or being
 killed by a shark?
Which is more likely: dying from stomach
 cancer or dying in a car accident?
 Cancer Screening
A doctor has just examined a woman for breast
 cancer. She has a lump in her breast, but based on
 years of experience the doctor estimates the odds
 of malignancy at 1 in 100. A mammogram is
 ordered. In general these tests accurately classify
 80% of malignant tumours and 90% of benign
 tumours.
If the mammogram says the tumour is malignant,
 what would you say the overall chances of the
 tumour being malignant are now?
Estimating probabilities:
sources of bias
Representativeness           Conservatism
Availability                 Risk perceptions
Confusion of the Inverse     Anchoring
Looking on the Bright Side   Coincidences
Compound Events

For more details see “The Psychology of Judgement
 and Decision Making” by Plous
Report back on homework
How closely did the probability estimates
 agree?
Why do you think they differed?
How would you go about coming up with a
 single number?
Using pictures to understand
probabilities
Venn Diagrams
  – great way to visualise simple probabilities and
    overlapping events
Probability Trees
  – useful for sequential events, repeated events,
    multiple items or people
Contingency Tables
  – useful for understanding, visualising and
    computing conditional probabilities
Venn Diagrams
Venn diagrams are used to visually convey the
 concept of sample spaces and events.
A rectangle is used to represent the sample
 space of an experiment and as such contains all
 possible sample points.
Circles within the rectangle represent events.
The area within the circle contains the sample
 points which belong to that event.
Simple Venn Diagram



       A        Ac
Venn Diagram

        A             B

   A  Bc     A B    Ac  B




            Ac  Bc
Venn Diagram of Union of
Events A and B



             AB



            Ac  Bc
Share price problem
P(A up) = 0.7,     P(B up) = 0.5
P(C up) = 0.3, P(A up  B up) = 0.3
P(A up  C up) = 0.1,
P(B up  C up) = 0.2
P(A up  B up  C up) = 0.05
What is P(A up  B up) ?
What is P(A up  B up  C up) ?
What is P(A up  B down  C down) ?
 Probability Trees
     Each node is an experiment. Each branch is
      an outcome. Branches that come out of a
      node should be exhaustive and mutually
                                          Conditional
      exclusive.                          probabilities
                           Prob= 0.7
         Prob= 0.6                      YY 0.42
                      Y
                     0.6   Prob= 0.3
                                        YN 0.18
 1                         Prob= 0.4
         Prob= 0.4                      NY 0.16
                      N
                     0.4   Prob= 0.6
                                        NN 0.24
Job offers
 Joint Probability Table

                         Event
                 A                 Ac

        B    P(A  B)            P(Ac  B)    P(B)
Event
        Bc   P(A  Bc)           P(Ac  Bc)   P(Bc)

                P(A)               P(Ac)       1
Convert Cross-Tabs Table Into
a Joint Probability Table
                     No JS    JS

Cross-Tabs    LOW
                      13      4
              PROD
Table
 (n = 36)     HIGH
              PROD    1      18
                                    Total = 36

                     No JS    JS

Joint         LOW
                     .361    .111     4/36
              PROD
Probability
Table         HIGH
              PROD   .028    .500
Joint and Union Probabilities
                  No JS     JS
          LOW
          PROD     .361    .111    .472
          HIGH
          PROD     .028    .500    .528
                     .389      .611 1.00
Unconditional Probability
      P(LOW PROD)          = .472
Joint Probability
      P(LOW PROD  JS) = .111
Union Probability
      P(LOW PROD  JS) = .361 + .111 + .500 = .972
Excel - Probability

             Pivot-tables
             Downloadable spreadsheet
              from webpage with
              “proforma” probability tables
Scholarships
In a study of 100 students who had been
 awarded university scholarships, it was
 found that 40 had part-time jobs, 25 had
 achieved honours in the previous semester,
 and 15 had both a part-time job and had
 achieved honours.
What was the probability that a student had
 a part-time job or had achieved honours (ie.
 the union probability)?
Expected Value
What is an expected value?
Notation: expected value of x is E(x)
Calculating E(x):

E ( x)  P( x1 ) x1  P( x2 ) x2  P( x3 ) x3  ...
  Expected Value example
If you want to rent out a house and you know:
 Weekly Rent          Probability
 $210                 0.4
 $225                 0.3
 $240                 0.25
 $250                 0.05

What do you expect to earn in rent each week?
Watch this space (Video)
Use probability language to describe the
 two main positions in this argument
Which side do agree with? Why?
Can you think of other examples where
 there are similar differences in interpreting
 probabilities and expected values?
Monty Hall/Let’s Make a Deal




 Door A    Door B     Door C
Conditional Probability
P(A|B) is the probability of A happening
 given that B has happened.
This is called a conditional probability
Important formula: (Bayes’ Theorem)
    P(A|B) = P(A  B) / P(B)
Or round the other way:
    P(A  B) = P(A|B) x P(B)
  Unconditional and Conditional
  Probabilities
                    No JS      JS
            LOW
                     .361     .111    .472
            PROD
            HIGH
            PROD
                     .028    .500     .528

                     .389     .611
Unconditional Probability
    P(HIGH PROD)           = .528
Conditional Probability
    P(HIGH PROD | JS) = .500/.611 = .81
    Only Interested in JS Groups. Thus .611 is the
    Denominator. High Prod is .500, the Numerator.
  Independent Events

Events are independent if the outcome of one
 does not affect the outcome of the other.
Example:
  – the profit made by Telstra is independent of what
    I chose to have for breakfast yesterday.
  – But the profit made by Telstra is not independent
    of whether I chose to phone my friend in
    California yesterday.
Statistical Independence ?
If P(A) = P(A | B), then Events A and B
 are Independent.
  – Does P(High Prod) = P(High Prod | Job
    Switch)?

Knowing Event B Happened Does Not
 Affect the Probability of A.
  Is Job Switching Related to
  Productivity Level?
   P(HIGH PROD)       = .528
   P(HIGH PROD | JS) = .500/.611 = .818
   P(HIGH PROD | NJS) = .028/.389 = .072
     Note: Three Probabilities are Different.
 Probability of High Productivity if Group Members
  Switch Jobs is .818.
 Probability of High Productivity if Group Members
  Do Not Switch Jobs is Only .072.
 Conclusion: High Productivity and Job Switching May
  be Related.
Probability Trees – independent
events
Trees can also be used when events are
 independent.
Probabilities on branches are just simple
 probabilities of each event occurring
This is because literal (symbolic) definition
 of independence is: P(A) = P(A | B)
Conditional probability is simple
 probability for independent events
  Mutually Exclusive Events

Events are mutually exclusive if they both
 couldn’t happen at once.
Example:
  – Telstra making a profit of more than $300 million is
    mutually exclusive with them making a profit of less
    than $300 million this year.
  – Me having toast for breakfast is not mutually
    exclusive with me having Weetbix for breakfast (I
    might have both).
Calculating Union probabilities
In general the probability of A  B
 happening is:
 P(A  B) = P(A) + P(B) - P(A  B)

If events are mutually exclusive then the
 probability of A  B happening is:
 P(A  B) = P(A) + P(B)
Calculating Intersection
probabilities
In general the probability of A  B
 happening must be found by gathering data
 to estimate it separately.

However if events are independent then the
 probability of A and B happening is:
 P(A  B) = P(A)  P(B)
Murder He Wrote
Do you understand the probabilities that are
 talked about in the article?

What does the law mean when it talks about
 “beyond a reasonable doubt”? Can you
 assign a probability to this?
The Birthday Problem
                                                              Birthday Problem

                         1
                        0.9
                        0.8
P(at least one match)




                        0.7
                        0.6
                        0.5
                        0.4
                        0.3
                        0.2
                        0.1
                         0
                                          10
                                               13

                                                    16
                                                         19

                                                               22
                                                                     25
                                                                          28

                                                                               31
                                                                                    34

                                                                                         37
                                                                                              40
                                                                                                   43

                                                                                                        46
                                                                                                             49

                                                                                                                  52
                                                                                                                       55
                                                                                                                            58
                              1
                                  4

                                      7




                                                                    Number of People (n)
Birthday Problem (the
calculations)
Probability of 2 people not having the same
 birthday = 364/365
Probability of 3 people not having the same
 birthday = Probability of 2 people not
 having the same birthday AND a 3rd person
 not having the same birthday as either of the
 first 2 = 364/365  363/365
Probability of 4 people not having the same
 birthday = 364/365  363/365  362/365
 Birthday problem (cont.)
Prob of no birthday matches with (n+1)
 people =        364  363 362  ... (365  n)
                              365n
Prob of (n+1) people having at least one
 matching birthday = 1 - P(no matches)
Example: P(no matches for 10 people) = 0.8831
P(at least one match for 10 people) = 0.1169
Wallet Problem

                 What is the
   5    5        probability that given
                 we select a five dollar
                 bill from one of these
    5   10       three wallets at
                 random that the other
                 bill in that wallet is a
   10   10       five?
 Probability Trees
Remember we used probability trees with
 conditional probabilities on the branches.
 Useful to depict sequential events.
                          P(C|A)=0.4
                                       AC 0.08
         P(A)=0.2    A
                    0.2   P(D|A)=0.6
                                       AD 0.12
    1                     P(C|B)=0.3
                                       BC 0.24
         P(B)=0.8    B
                    0.8   P(D|B)=0.7
                                       BD 0.56
Conditional probabilities
Can also think of a conditional probability as a
 probability of something happening within a
 “subset” of the population.
Use a Venn diagram or a probability table e.g.

                    P(A | B) = 10 / 50 = 0.2
      A        B
     20   10   40   (Notice that this is identical to
30
                    Bayes theorem formula)
Another way of thinking about
conditional probabilities
Many times during probability analysis, we
 may want to revise some prior probabilities
 P(A) based on new information (B).
Bayes Theorem allows us to do this. The
 new probabilities are called posterior
 probabilities P(A|B) and, as the new
 estimates of the likelihood of A, can be used
 in future calculations instead of P(A).
Excel – Bayes Theorem

           Download Bayes Theorem
            spreadsheet from webpage to
            do the calculations
Important Formulas

P ( A)  P ( A )  1
              c


P ( A  B )  P ( A  B )  P ( A)
                         c


P ( A  B )  P ( A)  P ( B )  P ( A  B )
P ( A B )  P ( B  A)       P( B)
P( A  B)  P( B) P( A B)
Rashes
A pharmaceutical company conducted a study to
 evaluate the effect of an allergy relief medicine; 250
 patients with symptoms that included itchy eyes and
 a skin rash were given a new drug. 90 of the patients
 experienced eye relief, 135 had their rash clear up,
 and 45 had relief from both conditions.
What is the likelihood that a patient will experience
 relief from at least one of these conditions?
Do these events appear independent? Explain.
Rush Orders
Rush orders for a raw material are placed with two
 different suppliers, A and B. If neither order arrives
 in 4 days production is shut down until at least one
 comes. P(A will deliver in 4 days) = 0.55,
 P(B will deliver in 4 days) = 0.35
P(both deliver in 4 days)?
P(at least one delivers in 4 days)?
P(production is shut down)?
Advertising Soap
B = bought, S = remembers seeing ad
P(B) = 0.2, P(S) = 0.4, P(B  S) = 0.12
P(B|S)? Does seeing the ad increase probability of
 purchasing? Would you keep advertising?
If those who don’t buy from you, buy from your
 competitor:
What would be your estimate of your market
 share?
Would your continuing the ad increase your
 market share? Why or why not?
Advertising Soap (cont.)
Another ad has been tested and has values of
 P(S)=0.3 and P(B  S) = 0.1.
What is P(B|S) for this ad?
Which ad has a bigger effect on purchases?
  Credit?
A bank is reviewing its credit card policy with a view
 to recalling some of its credit cards. In the past
 approximately 5% of cardholders have defaulted and
 the bank has been unable to collect the outstanding
 balance. The bank has further found that the
 probability of missing one or more payments for those
 customers who do not default is 0.20. Of course the
 probability of missing one or more payments for those
 who default is 1.
Given that a customer has missed a payment, compute
 the probability that they will default.
 Integrated Circuit chips
A process produces IC chips. Over the long-run
 the fraction of bad chips produced by the process
 is around 20%. Thorough testing is expensive but
 there is a cheap test. All good chips pass the cheap
 test but so do 10% of bad chips.
Given that a chip passes the test what is the
 probability that it is a good chip?
If a company sold all chips that passed the test
 what fraction of what they sold would be bad?
  Cancer Screening
Use Bayes Theorem to answer the cancer
 screening question from beginning of lecture.
Prior probability of malignancy = 0.01
P(positive test | malignant) = 0.8
P(negative test | benign) = 0.9
What is the posterior probability of malignancy?
 (I.e. the new probability of malignancy if you get
 information that the test was positive)
What did we do?
Discussed where intuition about probability fails
Discussed how probability is estimated
Venn Diagrams
Explored the use of tree diagrams and probability
 tables in calculating probability.
Talked about independence and mutual exclusivity
Expected value
Conditional probability
Talked about statistical independence
Worked on problems
Excel functions today
Expected values using =SUMPRODUCT()
Bayes theorem and probability tables
 spreadsheet – download and use to
 automate simple calculations. Practice using
 this to solve problems you get in class and
 in textbook/webpage.
Managerial applications
What did you learn today that makes a
 difference to the way you manage?
What are the three most important things to
 remember from today’s lecture?
Next class
Prepare case study “Who is the customer?”
 (Specific questions given on the class
 webpage). This is assessed if your syndicate
 chose option 1
Read and answer questions for “Hair Tonic”
Read supplementary readings on Decision
 Trees and Valuing Information
Answer additional questions on webpage

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:6
posted:2/3/2012
language:English
pages:57