Download Probability in our Daily Lives

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Learning Objectives
Probability in our Daily
Lives
1.  Random Phenomena
2.  Law of Large Numbers
3.  Probability
4.  Independent Trials
5.  Finding probabilities
Section 1: How can Probability
Quantify Randomness?
6.  Types of Probabilities: Relative Frequency
and Subjective
Learning Objective 1:
Random Phenomena
Learning Objective 2:
Law of Large Numbers
!  For random phenomena, the outcome is
!  As the number of trials increase, the proportion of
uncertain
!  In
the short-run, the proportion of times that
something happens is highly random
!  In the long-run, the proportion of times that
something happens becomes very predictable
occurrences of any given outcome approaches a
particular number “in the long run”
!  For example, as one tosses a fair die, in the long run
1/6 of the observations will be a 6.
Probability quantifies long-run randomness
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Learning Objective 3:
Probability
Learning Objective 4:
Independent Trials
!  Frequency Interpretation: With random
!  Different trials of a random phenomenon are
phenomena, the probability of a particular
outcome is the proportion of times that the
outcome would occur in a long run of
observations
!  Example:
independent if the outcome of any one trial is
not affected by the outcome of any other trial.
!  Example:
!  If
you have 20 flips of a coin in a row that are
“heads”, you are not “due” a “tail” - the
probability of a tail on your next flip is still 1/2.
The trial of flipping a coin is independent of
previous flips.
!  When
rolling a fair die, the outcome of “6” has
probability = 1/6. In other words, the
proportion of times that a 6 would occur in a
long run of observations is 1/6.
Learning Objective 5:
How can we find Probabilities?
Learning Objective 6:
Types of Probability: Relative Frequency vs.
Subjective
!  Calculate theoretical probabilities based on
!  The relative frequency definition of probability is the
assumptions about the random phenomena.
For example, it is often reasonable to assume
that outcomes are equally likely such as when
flipping a coin, or a rolling a die.
!  Observe many trials of the random
phenomenon and use the sample proportion,
i.e., the relative frequency of the number of
times the outcome occurs as its probability.
This is merely an estimate of the actual
probability.
long run proportion of times that the outcome occurs
in a very large number of trials - not always helpful/
possible.
!  When a long run of trials is not feasible, you must rely
on subjective information. In this case, the subjective
definition of the probability of an outcome is your
degree of belief that the outcome will occur based on
the information available.
! 
Bayesian statistics is a branch of statistics that uses
subjective probability as its foundation
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Learning Objectives
Probability in Our Daily
Lives
1.  Sample Space
2.  Event
3.  Probabilities for a sample space
4.  Probability of an event
5.  Basic rules for finding probabilities about a
pair of events
Section 2: How Can We Find Probabilities?
Learning Objectives
Learning Objective 1:
Sample Space
6.  Probability of the union of two events
!  For a random phenomenon, the sample
7.  Probability of the intersection of two events
space is the set of all possible outcomes.
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Learning Objective 2:
Event
Learning Objective 3:
Probabilities for a sample space
!  An event is a subset of the sample space
Each outcome in a sample space has a
probability
!  The probability of each individual outcome is
between 0 and 1
!  The total of all the individual probabilities
equals 1.
!  An event corresponds to a particular outcome
or a group of possible outcomes.
!  For example;
!  Event A = student answers all 3 questions
correctly = (CCC)
!  Event B = student passes (at least 2 correct) =
(CCI, CIC, ICC, CCC)
Learning Objective 4:
Probability of an Event
Learning Objective 4:
Example: What are the Chances of being Audited?
!  The probability of an event A, denoted by
P(A), is obtained by adding the probabilities
of the individual outcomes in the event.
!  When all the possible outcomes are equally
likely:
!  What is the sample space for selecting a
number of outcomes in event A
P ( A) =
number of outcomes in the sample space
taxpayer?
{(under $25,000, Yes), (under $25,000, No),
($25,000 - $49,000, Yes) …}
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Learning Objective 4:
Example: What are the Chances of being Audited?
For a randomly selected taxpayer in 2002,
!  What is the probability of an audit?
!  310/80200=0.004
!  What is the probability of an income of
$100,000 or more?
!  10700/80200=0.133
!  What income level has the greatest
probability of being audited?
!  $100,000
Learning Objective 5:
Basic rules for finding probabilities about a pair
of events
!  Some events are expressed as the outcomes
that
!  Are
not in some other event (complement of
the event)
!  Are in both one event and in another event
(intersection of two events)
!  Are in either one event or in another event – or
both (union of two events)
or more = 80/10700= 0.007
Learning Objective 5:
Complement of an event
Learning Objective 5:
Disjoint events
!  The complement of an event A consists of all
!  Two events, A and B, are disjoint or mutually
outcomes in the sample space that are not in
A.
!  The probabilities of A and of Ac add to 1
!  P(Ac) = 1 – P(A)
exclusive if they do not have any common
outcomes
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Learning Objective 5:
Intersection of two events
Learning Objective 5:
Union of two events
!  The intersection of A and B consists of
!  The union of A with B consists of outcomes
outcomes that are in both A and B
that are either in A or in B or in both A and B.
Learning Objective 6:
Probability of the Union of Two Events
Learning Objective 6:
Example
Addition Rule:
For the union of two events,
P(A or B) = P(A) + P(B) – P(A and B)
!  80.2 million tax payers (80,200 thousand)
If the events are disjoint, P(A and B) = 0, so
P(A or B) = P(A) + P(B)
!  Event A = being audited
!  Event B = income greater than $100,000
!  P(A and B) = 80/80200=.001
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Learning Objective 7:
Probability of the Intersection of Two Events
Learning Objective 7:
Example
Multiplication Rule for Independent Events:
!  What is the probability of getting 3 questions
For the intersection of two independent events,
A and B,
P(A and B) = P(A) x P(B)
correct by guessing?
!  Probability of guessing correctly is .2
!  This is a definition of probabilistic
independence
What is the probability that a student
answers at least 2 questions correctly?
!  We will give a justification for this later
P(CCC) + P(CCI) + P(CIC) + P(ICC) =
0.008 + 3(0.032) = 0.104
Learning Objective 7:
Assuming independence
Learning Objective 7:
Events Often Are Not Independent
!  Don’t assume that events are independent
!  Example: A Pop Quiz with 2 Multiple Choice
unless you have given this assumption
careful thought and it seems plausible.
Questions
!  Data
giving the proportions for the actual
responses of students in a class
Outcome: II
IC
Probability: 0.26 0.11
CI
CC
0.05
0.58
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Learning Objective 7:
Events Often Are Not Independent
!  Define the events A and B as follows:
A: {first question is answered correctly}
B: {second question is answered correctly}
!  P(A) = P{(CI), (CC)} = 0.05 + 0.58 = 0.63
!  P(B) = P{(IC), (CC)} = 0.11 + 0.58 = 0.69
!  P(A and B) = P{(CC)} = 0.58
! 
! 
!  If A and B were independent,
P(A and B) = P(A) x P(B) = 0.63 x 0.69 = 0.43
Thus, in this case, A and B are not independent!
Probability in Our Daily
Lives
Section 3: Conditional Probability:
What’s the Probability of A, Given B?
Learning Objectives
Learning Objective 1:
Conditional Probability
1.  Conditional probability
!  For events A and B, the conditional probability of
2.  Multiplication rule for finding P(A and B)
3.  Independent events defined using
conditional probability
event A, given that event B has occurred, is:
P(A | B) =
P(A & B)
P(B)
!  P(A|B) is read as “the probability of event A, given
event B.” The vertical slash represents the word
“given”. Of the times that B occurs, P(A|B) is the
proportion of times that A also occurs
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Learning Objective 1:
Conditional Probability
Learning Objective 1:
Example 1
Learning Objective 1:
Example 1
Learning Objective 1:
Example 1
!  What was the probability of being audited,
given that the income was ≥ $100,000?
!  Event
!  Event
A: Taxpayer is audited
B: Taxpayer’s income ≥ $100,000
P(A | B) =
P(A and B) 0.0010
=
= 0.007
P(B)
0.1334
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Learning Objective 1:
Example 1
Learning Objective 1:
Example 2
!  What is the probability of being audited given that the
!  A study of 5282 women aged 35 or over
income level is < $25,000
!  Let A =Event being Audited
!  Let B = Income < $25,000
P(A | B) =
analyzed the Triple Blood Test to test its
accuracy
P(A & B)
P(B)
!  P(A and B) = .0011
!  P(B)=.1758
!  .0011/.1758=.0063
Learning Objective 1:
Example 2
!  A positive test result states that the
condition is present
!  A negative test result states that the
condition is not present
Learning Objective 1:
Example 2
!  Assuming the sample is representative of the
population, find the estimated probability of a
positive test for a randomly chosen pregnant
woman 35 years or older
!  P(POS) = 1355/5282 = 0.257
!  False Positive: Test states the condition
is present, but it is actually absent
!  False Negative: Test states the condition
is absent, but it is actually present
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Learning Objective 1:
Example 2
Learning Objective 2:
Multiplication Rule for Finding P(A and B)
!  Given that the diagnostic test result is positive,
find the estimated probability that Down
syndrome truly is present
P(D and POS)
48 / 5282
P(D | POS) =
=
=
P(POS)
1355 / 5282
0.009
= 0.035
0.257
!  For events A and B, the probability that A
and B both occur equals:
!  P(A
and B) = P(A|B) x P(B)
also
!  P(A and B) = P(B|A) x P(A)
!  Summary: Of the women who tested positive,
fewer than 4% actually had fetuses with Down
syndrome
Learning Objective 2:
Example
Learning Objective 2:
Example
!  Roger Federer – 2006 men’s champion in
!  Assuming these are typical of his serving
the Wimbledon tennis tournament
!  He
made 56% of his first serves
!  He faulted on the first serve 44% of the
time
!  Given that he made a fault with his first
serve, he made a fault on his second serve
only 2% of the time
performance, when he serves, what is the
probability that he makes a double fault?
!  P(F1) = 0.44
!  P(F2|F1) = 0.02
!  P(F1 and F2) = P(F2|F1) x P(F1)
= 0.02 x 0.44 = 0.009
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Learning Objective 3:
Independent Events Defined Using Conditional
Probabilities
Learning Objective 3:
Checking for Independence
!  To determine whether events A and B are
!  Two events A and B are independent if the
probability that one occurs is not affected
by whether or not the other event occurs
!  Events A and B are independent if:
P(A|B) = P(A), or equivalently, P(B|A) = P(B)
independent:
!  Is
P(A|B) = P(A)?
P(B|A) = P(B)?
!  Is P(A and B) = P(A) x P(B)?
!  Is
!  If any of these is true, the others are also true and
the events A and B are independent
!  Events A and B are independent if and
only if
P(A and B) = P(A) x P(B)
Learning Objective 3:
Example
Learning Objective 3:
Example: Checking for Independent
!  The diagnostic blood test for Down syndrome:
Status
!  Are the events POS and D independent or
dependent? Is P(POS|D) = P(POS)?
POS = positive result
NEG = negative result
D = Down Syndrome
DC = Unaffected
!  P(POS|D) =P(POS and D)/P(D)
POS
NEG
Total
D
0.009
0.001
0.010
Dc
0.247
0.742
0.990
Total
0.257
0.743
1.000
= 0.009/0.010 = 0.90
!  P(POS) = 0.257
!  The events POS and D are dependent
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Learning Objectives
Probability in Our Daily
Lives
1.  Is a “Coincidence” Truly an Unusual Event?
2.  Probability Model
3.  Probabilities and Diagnostic Testing
4.  Simulation
Section 4: Applying the Probability Rules
Learning Objective 1:
Is a “Coincidence” Truly an Unusual Event?
!  The law of very large numbers states that
if something has a very large number of
opportunities to happen, occasionally it
will happen, even if it seems highly
unusual
Learning Objective 1:
Example: Is a Matching Birthday Surprising?
!  What is the probability that at least two
students in a group of 25 students have
the same birthday?
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Learning Objective 1:
Example: Is a Matching Birthday Surprising?
!  P(at least one match) = 1 – P(no matches)
Learning Objective 1:
Example: Is a Matching Birthday Surprising?
!  P(no matches) = P(students 1 and 2 and 3
…and 25 have different birthdays)
Learning Objective 1:
Example: Is a Matching Birthday Surprising?
Learning Objective 1:
Example: Is a Matching Birthday Surprising?
!  P(no matches) =
(365/365) x (364/365) x (363/365) x …
x (341/365)
!  P(no matches) = 0.43
!  P(at least one match) =
1 – P(no matches) = 1 – 0.43 = 0.57
Not so surprising when you consider that there
are 300 pairs of students who can share the
same birthday!
(This is the number of subsets of size 2 from the set of size 25)
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Learning Objective 2:
Probability Model
Learning Objective 2:
Probability Model
!  We’ve dealt with finding probabilities in many
!  A probability model specifies the possible
idealized situations
!  In practice, it’s difficult to tell when outcomes
are equally likely or events are independent
!  In most cases, we must specify a probability
model that approximates reality
outcomes for a sample space and provides
assumptions on which the probabilities for
events composed of these outcomes are
based
!  Probability models merely approximate reality
Learning Objective 2:
Example: Probability Model
Learning Objective 2:
Example: Probability Model
!  Out of the first 113 space shuttle missions
!  This answer relies on the assumptions of
there were two failures
!  What is the probability of at least one failure
in a total of 100 missions?
!  P(at
least 1 failure)=1-P(0 failures)
=1-P(S1 and S2 and S3 … and S100)
=1-P(S1)xP(S2)x…xP(S100)
=1-[P(S)]100=1-[0.971]100=0.947
! 
! 
Same success probability on each flight
Independence
These assumptions are suspect since other
variables (temperature at launch, crew
experience, age of craft, etc.) could affect the
probability
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Learning Objective 3:
Probabilities and Diagnostic Testing
Learning Objective 3:
Example: Probabilities and Diagnostic Testing
Random Drug Testing of Air Traffic
Controllers
!  Sensitivity of test = 0.96
!  Specificity of test = 0.93
!  Probability of drug use at a given time ≈
0.007 (prevalence of the drug)
!  Sensitivity = P(POS|S)
!  Specificity = P(NEG|SC)
Learning Objective 3:
Example: Probabilities and Diagnostic Testing
Learning Objective 4:
Simulation
What is the probability of a positive test result?
Some probabilities are very difficult to find
with ordinary reasoning. In such cases,
we can approximate an answer by
simulation.
P(POS)=P(S and POS)+P(SC and POS)
!  P(S and POS)=P(S)P(POS|S)
= 0.007x0.96=0.0067
! 
P(SC and POS)=P(SC)P(POS|SC)
! 
P(POS)=.0067+.0695=0.0762
= 0.993x0.07=0.0695
Even though the prevalence is < 1%, there is an almost
8% chance of the test suggesting drug use!
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Learning Objective 4:
Simulation
Carrying out a Simulation:
Combinatorics
"How do I love thee"by Elizabeth Barrett Browning (1806-1861)
How do I love thee? Let me count the ways.
I love thee to the depth and breadth and height
My soul can reach, when feeling out of sight
For the ends of Being and ideal Grace.
I love thee to the level of everyday's
Most quiet need, by sun and candle-light.
I love thee freely, as men strive for Right;
I love thee purely, as they turn from Praise.
I love thee with a passion put to use
In my old griefs, and with my childhood's faith.
I love thee with a love I seemed to lose
With my lost saints, --- I love thee with the breath,
Smiles, tears, of all my life! --- and, if God choose,
I shall but love thee better after death.
!  Identify
the random phenomenon to be
simulated
!  Describe how to simulate observations
!  Carry out the simulation many times (at least
1000 times)
!  Summarize results and state the conclusion
Addition Principle
The number of items in the union of
mutually disjoint sets is the sum of the
number in each set.
Multiplication Principle
If a procedure involves a sequence of
choices, in such a way that the number of
choices at each stage is independent of
the specific choices in prior stages, then
the total number of outcomes is the
product of the outcomes in each stage.
Permutations
The number of permutations of n objects,
taken k at a time, is
P(n,k) =
n(n −1)(n − 2)...2.1
=
n!
(n − k)!
Combinations
The number of ways of selecting a subset
of k objects from a set of n objects is given
by
C(n,k) =
n(n −1)(n − 2)...2.1
n!
=
k(k −1)...2.1
k!(n − k)!
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