Transcript Review of Probability

Quick Review of Probability
Event
state of the world that may or may
not occur --- A, B, C, etc.
Joint Event
events occurring together --- “A and
B”, AB = BA
Conditional Event
an event occurring given that
another occurs --- “A given B”, A|B
--- “B given A”, B|A
Mut. Exclusive Events
events that cannot occur together
Coll. Exhaustive Events
events that together comprise all
possibilities
Independent Events
events that do not depend on one
another
P(A)
“single-event” probability
P(AB) = P(BA)
joint probability
P(A|B), P(B|A)
conditional probability
IMPORTANT FACT
If several events are mutually exclusive
and collectively exhaustive, then their
probabilities add to 1.0
“The MECE rule”
General Formulas
Conditional Probability
P(A|B) = P(AB) / P(B)
P(B|A) = P(BA) / P(A)
Multiplication Rule
P(AB) = P(A|B) P(B) = P(B|A) P(A)
Special Formulas for Independent Events
If you know that A and B are independent…
Conditional Probability
P(A|B) = P(A)
P(B|A) = P(B)
Multiplication Rule
P(AB) = P(A) P(B)
Do not use special formulas unless you
know A and B are independent!
Probability Trees
A probability tree is a graphical way to visualize the
possible outcomes of a sequence of occurrences
P(A)
mutually exclusive,
D P(D|A)
collectively exhaustive
E P(E|A)
A
F P(F|A)
G P(G|B)
B
H P(H|B)
P(B)
mutually exclusive,
collectively exhaustive
P(A) + P(B) = 1.0
mutually exclusive,
collectively exhaustive
P(D|A) + P(E|A) + P(F|A) = 1.0
P(G|B) + P(H|B) = 1.0
Information on a Probability Tree
There are two types of probabilities on a probability tree:
• “Single event” probabilities
• Conditional probabilities
Joint probabilities are not on the
tree, but they are easy to calculate
from the tree. For example:
P(AD) = P(D|A) P(A)
P(A)
D P(D|A)
A
E P(E|A)
B
P(B)
G P(G|B)
H P(H|B)
F P(F|A)
Tips for Probability Problems
Always be aware of:
• the events (single, joint, or conditional?)
• which probabilities are given
• which probabilities are asked for
• which special situations exist (independence?)
(may be given, or you may have to deduce)
Bayes’ Rule and “Too Many Bugs”
Public health scientists estimate that 1% of the general
population regularly abuses illegal substances. As the president
of a 2,000-employee company, you would like to use drug
testing to eliminate this 1% from your work force.
A new drug test has been devised that is relatively inexpensive
and that the manufacturer claims has a 95% accuracy rate. Your
idea is to test each employee, terminating or keeping the
employee based on the results of the test.
Is this a reasonable and fair plan?
D = event that a person uses Drugs
F = event that a person is drug Free
A = event that the test Accuses a person
E = event that the test Exonerates a person
What does “95% accuracy” mean?
This is terminology for how often the test confirms the truth.
P(A | D) = 0.95 and P(E | F) = 0.95
0.01
D
A 0.95
E 0.05
A 0.05
F
0.99
E 0.95
What are the probabilities of these events?
• Person is accused correctly
= AD
• Person is accused falsely
= AF
• Person is exonerated correctly
= EF
• Person is exonerated falsely
= ED
Using the multiplication rule (events not independent):
•
•
•
•
P(AD) = P(A | D) P(D) = 0.95 * 0.01
P(AF) = P(A | F) P(F) = 0.05 * 0.99
P(EF) = P(E | F) P(F) = 0.95 * 0.99
P(ED) = P(E | D) P(D) = 0.05 * 0.01
Have 4.95% accused falsely,
but only 1% actually use drugs!
= 0.0095
= 0.0495
= 0.9405
= 0.0005
What are P(A) and P(E)? Because mut. exclusive:
• P(A) = P(AD) + P(AF) = 0.0095 + 0.0495 = 0.0590
• P(E) = P(EF) + P(ED) = 0.9405 + 0.0005 = 0.9410
What are P(D | A) and P(F | E)? By the conditional
probability rule (events not independent):
• P(D | A) = P(AD) / P(A) = 0.0095 / 0.0590 = 0.1610
• P(F | E) = P(EF) / P(E) = 0.9405 / 0.9410
What does this mean exactly?
= 0.9995
Consider the following situation. Suppose John Smith has
been accused by the drug test. You have no way of knowing
for sure if John uses drugs. Do you fire John based on the test
alone?
P( John uses | John has been accused ) = P( D | A) = 0.1610
There is a 16% chance John abuses drugs
even though the test has accused him
Now suppose John Smith has been exonerated by the test. Do you
keep John as an employee?
P( John is free | John has been exonerated ) = P( F | E) = 0.9995
It’s 99.95% sure that John does not use
drugs based on the result of his test
The moral of the “Bugs” article:
Even if a person is accused by a test, the conditional
probabilities show that there is a good chance that the
person has been falsely accused.
On the other hand, if a person is exonerated by
a test, then it is nearly certain that the person
deserves to be exonerated.
The process we went through is Bayes’ Rule, and it
can be used in several additional ways (not just in
“drug testing” scenarios)
Test Marketing a New Product
New products introduced in the marketplace have high sales
8% of the time and low sales 92% of the time. A marketing
test has the following accuracies: if sales are high, then
consumer test reaction is positive 70%, neutral 25%, and
negative 5%; if sales are low, then consumer test reaction is
positive 15%, neutral 35%, and negative 50%.
Your company is developing a new product and will be test
marketing to better gauge the sales of the new product. Based
on positive, neutral, or negative reactions, what are the
probabilities of high and low sales?
Decision Analysis
We will now be using probability trees to help
us make decisions in the face of uncertainty
Ingredients for a quantitative decision
• Possible decisions
• Uncertain events
• Payoffs or costs to decisions and events
• Probabilities of the events
One day each weekend, you rent an indoor booth to sell your homemade crafts at
the J&J Flea Market. From your experience, you know the following:
Profit/Weather
No Rain
Rain
Saturday
$1000
$500
Sunday
$700
$350
This Saturday, there is a 70% chance of rain, and this Sunday, there is a 30%
chance. Which day do you sell at the flea market?
Actions: sell on Saturday,
sell on Sunday
Payoffs: dollar amounts
Events: rain or not on
Saturday, rain or not on
Sunday
Probs: 70%, 30%, etc.
How to Evaluate a Decision
There are two stages to evaluating a decision
•
FORWARD stage
Make a decision tree that lays out all possible sequences of
decisions and events, along with the payoffs and probs
•
BACKWARD stage
Evaluate EMVs or EMCs from the end to the beginning
What are EMVs and EMCs?
Will explain shortly…
Forward Stage for J&J
Sat
$0
0.7
Rain
$500
0.3
No Rain
$1000
0.3
Rain
$350
Sun
$0
0.7
No Rain
$700
EMVs and EMCs
EMV = Expected Monetary Value
= the average payoff of a series of future decisions
and events, assuming that, at any time during the
decision-making process, the decision-maker
makes the wisest decisions given the pending
uncertainties
EMC = Expected Monetary Cost
= the average cost of a series of future decisions
and events, assuming that, at any time during the
decision-making process, the decision-maker
makes the wisest decisions given the pending
uncertainties
Goal: maximize EMV and minimize EMC!
Backward Stage for J&J
$650
Sat
$0
$650
$595
Sun
$0
0.7
Rain $500
$500
0.3
No Rain $1000
$1000
0.3
Rain $350
$350
0.7
No Rain $700
$700
Rules for Evaluating EMVs
• The EMVs at the ends of the paths in the tree are
the total cumulative payoffs gotten over each
path
• The EMV of an event (circle) is the average of
the EMVs of the event’s branches, weighted by
the probs
• The EMV of a decision (square) is the maximum
of the EMVs of the decision’s branches
Rules for Evaluating EMCs
• The EMCs at the ends of the paths in the tree are
the total cumulative costs gotten over each path
• The EMC of an event (circle) is the average of
the EMCs of the event’s branches, weighted by
the probs
• The EMC of a decision (square) is the minimum
of the EMCs of the decision’s branches
EMVs and EMCs are similar except
“payoffs vs costs” and “maximize vs minimize”
Example for EMV of Event
P(A)
A
$X
EMV
P(B)
EMVA
EMVB
B
$Y
P(C)
C
$Z
EMVC
EMV = P(A)*EMVA + P(B)*EMVB + P(C)*EMVC
Note: payoffs $X, $Y, and $Z are not in the formula
Fruit Flies Discussion
(preparation for Homework 2)
• In 1981, Mediterranean fruit flies infested crops in Santa
Clara County, CA, threatening California business
• Spraying for the flies was an option, but spraying was
environmentally questionable
• Governor Brown and others dismissed spraying
• Instead, a control program of “fruit stripping” was begun,
depositing 750 tons of fruit in California landfills
• Then, the USDA came up with the “sterile male” solution,
attempting to capitalize on a biological trait of the flies
Fruit Flies Discussion
(preparation for Homework 2)
• Female fruit flies only mate once
• Radiation was used to sterilize a large quantity of male
fruit flies
• The males were then released to halt fly population growth
• Problem! The irradiation was done incorrectly, and the
males weren’t sterile
• More flies, more flies, and more flies
• Embargoes against California fruit
• Governor Brown decides to spray
• But it’s too late, his career is over