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Econ 240A
Power Three
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Summary: Week One
• Descriptive Statistics
– measures of central tendency
– measures of dispersion
• Distributions of observation values
– Histograms: frequency(number) Vs. value
• Exploratory data Analysis
– stem and leaf diagram
– box and whiskers diagram
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Probability
The Gambler
Kenny Rogers
20 Great Years
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Outline
• Why study probability?
• Random Experiments and Elementary
Outcomes
• Notion of a fair game
• Properties of probabilities
• Combining elementary outcomes into
events
• probability statements
• probability trees
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Outline continued
• conditional probability
• independence of two events
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Why study probability?
• Understand the concept behind a random
sample and why sampling is important
– independence of two or more events
• understand a Bernoulli event
– example; flipping a coin
• understand an experiment or a sequence of
independent Bernoulli trials
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Cont.
• Understand the derivation of the binomial
distribution, i.e. the distribution of the
number of successes, k, in n Bernoulli trials
• understand the normal distribution as a
continuous approximation to the discrete
binomial
• understand the likelihood function, i.e. the
probability of a random sample of
observations
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Concepts
• Random experiments
• Elementary outcomes
• example: flipping a coin is a random
experiment
– the elementary outcomes are heads, tails
• example: throwing a die is a random
experiment
– the elementary outcomes are one, two, three,
four, five, six
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Concept
• A fair game
• example: the probability of heads, p(h),
equals the probability of tails, p(t):
p(h) = p(t) =1/2
• example: the probability of any face of the
die is the same, p(one) = p(two) = p(three) =
p(four) =p(five) = p(six) = 1/6
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Uncertainty in Life
• Demography
– Death rates
– Marriage
– divorce
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Uncertainty in Life: US (CDC)
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Probability of First Marriage by Age, Women: US (CDC)
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Cohabitation: The Path to Marriage?: US(CDC)
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Race/ethnicity Affects Duration of First Marriage
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Properties of probabilities
• Nonnegative
0
– example:
p(h)
• probabilities of elementary events sum to
one
– example p(h) + p(t) = 1
Flipping a coin twice: 4
elementary outcomes
heads
h, h
tails
h, t
heads
heads t, h
tails
tails t, t
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Throwing Two Dice, 36
elementary outcomes
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Larry Gonick and
Woollcott Smith,
The Cartoon Guide
to Statistics
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Combining Elementary Outcomes
Into Events
• Example: throw two dice: event is white die
equals one
• example: throw two dice and red die equals
one
• example: throw two dice and the sum is
three
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Event: white die equals one is the bottom row
Event: red die equals one is the right hand column
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Event: 2 dice sum
to three is lower
diagonal
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Operations on events
• The event A and the event B both occur: ( A  B)
• Either the event A or the event B occurs or
both do: ( A  B)
• The event A does not occur, i.e.not A: A
Probability statements
• Probability of either event A or event B
p( A  B)  p( A)  p( B)  p( A  B)
– if the events are mutually exclusive, then p ( A  B)  0
• probability of event B
p ( B)  1  p ( B )
Probability of a white one or a red one: p(W1) + p(R1) double counts
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Two dice are thrown: probability of the white die showing
one and the red die showing one
p(W 1  R1)
Probability 2 dice
add to 6 or add to 3
are mutually
exclusive events
Probability of not
rolling snake eyes
is easier to calculate
as one minus the
probability of rolling
snake eyes
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Problem
• What is the probability of rolling at least one
six in two rolls of a single die?
– At least one six is one or two sixes
p(one6  two6' s)  1  p( zero 6' s)
– easier to calculate the probability of rolling zero
sixes: (5/36 + 5/36 + 5/36 + 5/36 + 5/36) = 25/36
– and then calculate the probability of rolling at
least one six: 1- 25/36 = 11/36
Probability tree
1
2
1
3
4
2
5
3
6
4
5
2 rolls of a die:
36 elementary
outcomes, of
which 11
involve one or
more sixes
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Conditional Probability
• Example: in rolling two dice, what is the
probability of getting a red one given that
you rolled a white one?
– P(R1/W1) ?
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In rolling two dice, what is the probability of getting a red one given
that you rolled a white one?
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Conditional Probability
• Example: in rolling two dice, what is the
probability of getting a red one given that
you rolled a white one?
– P(R1/W1) ?
p( R1 / W 1)  p( R1  W 1) / p(W 1)  (1 / 36) /(1 / 6)
Independence of two events
• p(A/B) = p(A)
– i.e. if event A is not conditional on event B
– then p A  B  p( A) * p(B)
Concept
• Bernoulli Trial
– two outcomes, e.g. success or failure
– successive independent trials
– probability of success is the same in each trial
• Example: flipping a coin multiple times
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Problem 6.28
cash
Credit card Debit card
0.09
0.03
0.04
$20-$100 0.05
0.21
0.18
>$100
0.23
0.14
<$20
0.03
Distribution of a retail store purchases classified by amount
and method of payment
Problem (Cont.)
• A. What proportion of purchases was paid
by debit card?
• B. Find the probability a credit card
purchase was over $100
• C. Determine the proportion of purchases
made by credit card or debit card
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Problem 6.28
cash
Credit card Debit card
0.09
0.03
0.04
$20-$100 0.05
0.21
0.18
>$100
0.03
0.23
0.14
Total
0.17
0.47
0.36
<$20
Problem (Cont.)
• A. What proportion of purchases was paid
by debit card? 0.36
• B. Find the probability a credit card
purchase was over $100
• C. Determine the proportion of purchases
made by credit card or debit card
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Problem (Cont.)
• A. What proportion of purchases was paid
by debit card?
• B. Find the probability a credit card
purchase was over $100 p(>$100/credit
card) = 0.23/0.47 = 0.489
• C. Determine the proportion of purchases
made by credit card or debit card
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Problem (Cont.)
• A. What proportion of purchases was paid by debit
card?
• B. Find the probability a credit card purchase was
over $100
• C. Determine the proportion of purchases made by
credit card or debit card
– note: credit card and debit card purchases are mutually
exclusive
– p(credit or debit) = p(credit) + p (debit) = 0.47 + 0.36
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Problem 6.61
• A survey of middle aged men reveals that
28% of them are balding at the crown of
their head. Moreover, it is known that such
men have an 18% probability of suffering a
heart attack in the next ten years. Men who
are not balding in this way have an 11%
probability of a heart attack. Find the
probability that a middle aged man will
suffer a heart attack in the next ten years.
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P (Bald and MA) = 0.28
Bald
Not Bald
Middle Aged men
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P (Bald and MA) = 0.28
P(HA/Bald and MA) = 0.18
Bald
P(HA/Not Bald and MA)
= 0.11
Not Bald
Middle Aged men
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Probability of a heart attack in
the next ten years
• P(HA) = P(HA and Bald and MA) + P(HA
and Not Bald and MA)
• P(HA) = P(HA/Bald and MA)*P(BALD
and MA) + P(HA/Not BALD and MA)*
P(Not Bald and MA)
• P(HA) = 0.18*0.28 + 0.11*0.72 = 0.054 +
.0792 = 0.1296
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Summary: Probability Rules
• Addition: P(A or B) = P(A) + P(B) – P(A and B)
– If A and B are mutually exclusive, P(A and B) = 0
• Subtraction: P(A) = 1 – P( not E)
• Multiplication: P(A and B) = P(A/B) P(B)
– If A and B are independent, then P(A/B) = P(B)
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