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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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Perspectives About Probability• Logical Discipline (like economics)

– Axiomatic: conclusions follow from assumptions

• Easier to Understand with Examples– I will use words, symbols and pictures

• Test Your Understanding By Working Problems

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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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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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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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Axiomatic Basis or Concepts

• Elementary outcomes have non-negative probabilities: P(H)>=0, P(T)>=0

• The sum of the probabilities over all elementary outcomes equals one: P(H) + P(T) = 1

HH

T

Flip a coin

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Axiomatic Basis or Concepts II• The probability of two mutually exculsive

events is zero: P(H and T) = P(H^T) = 0

• The probability of one outcome or the other is the sum of the probabilities of each minus any double counting: P(H or T) = P(H U T) = P(H) + P(T) – P(H^T)

• The probability of the event not happening is one minus the probability of the event happening: )(1)()( HPTPHP

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Axiomatic Basis or Concepts III

• Conditional probability of heads given tails equals the joint probability divided by the probability of tails: P(H/T) = P(H^T)/P(T)

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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

Properties of probabilities

• Nonnegative– example: p(h)

• probabilities of elementary events sum to one– example p(h) + p(t) = 1

0

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Another Example: Toss Two Coins

H1

T1

H2

T2

H, H

H, T

H2

T2

T, H

T, T

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Flipping a coin twice: 4 elementary outcomes

heads

tails

heads

tails

heads

tails

h, h

h, t

t, h

t, t

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Axiomatic Basis or Concepts

• Elementary outcomes have non-negative probabilities: P(H, H)>=0, P(H, T)>=0, P(T, H)>=0, P(T, T) >=0

• The sum of the probabilities over all elementary outcomes equals one: P(H, H) + P(H, T) + P(T, H) + P(T, T) = 1

• The probability of two mutually exculsive events is zero: P[(H, H)^(H, T)] = 0

H

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Axiomatic Basis or Concepts II• The probability of one outcome or the other

is the sum of the probabilities of each minus any double counting: P[(H, H) U (H,T)] = P(H, H) + P(H, T) – P[(H, H)^(H, T)] = P(H, H) + P(H, T)

• The probability of the event not happening is one minus the probability of the event happening:

),(),(),(),(1),( TTPHTPTHPHHPHHP

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Axiomatic Basis or Concepts III

• Conditional probability of heads, heads given heads, tails equals the joint probability divided by the probability of heads, tails: P[(H, H)/(H, T)] = P[(H, H)^(H, T)]/P(H, T)

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Throwing Two Dice, 36 elementary outcomes

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Larry Gonick and Woollcott Smith,The Cartoon Guideto 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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Combining Elementary Outcomes Into Events

• Example: throw two dice: event is white die equals one P(W1) =P(W1^R1) + P(W1^R2) + P(W1^R3) + P(W1^R4) + P(W1^R5) + P(W1^R6) = 6/36

• example: throw two dice and red die equals one

• example: throw two dice and the sum is three

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Event: 2 dice sum to three is lower diagonal

Operations on events

• The event A and the event B both occur:

• Either the event A or the event B occurs or both do:

• The event A does not occur, i.e.not A:

)( BA

)( BA

A

Probability statements

• Probability of either event A or event B

– if the events are mutually exclusive, then

• probability of event B

)()()()( BApBpApBAp

)(1)( BpBp

0)( BAp

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Probability of a white one or a red one: p(W1) + p(R1) double counts

Two dice are thrown: probability of the white die showing one and the red die showing one

36/1136/136/636/6)11(

)11()1()1()11(

36/1)11(

RWP

RWPRPWPRWP

so

RWP

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Probability 2 diceadd to 6 or add to 3 are mutually exclusive events

Probability of not rolling snake eyesis easier to calculateas one minus the probability of rolling snake eyes

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

– 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

)'6(1)'66( szeropstwoonep

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1

2

3

4

5

6

1

2

3

4

5

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Probability tree

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 giventhat you rolled a white one?

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) ?

)6/1/()36/1()1(/)11()1/1( WpWRpWRp

Independence of two events

• p(A/B) = p(A)– i.e. if event A is not conditional on event B– then )(*)( BpApBAp

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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

Problem 6.28

cash Credit card Debit card

<$20 0.09 0.03 0.04

$20-$100 0.05 0.21 0.18

>$100 0.03 0.23 0.14

Distribution of a retail store purchases classified by amountand method of payment

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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

Problem 6.28

cash Credit card Debit card

<$20 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

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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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Middle Aged men

Bald

P (Bald and MA) = 0.28

Not Bald

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Middle Aged men

Bald

P (Bald and MA) = 0.28

Not Bald

P(HA/Bald and MA) = 0.18

P(HA/Not Bald and MA)= 0.11

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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(E) = 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)