1

Econ 240A

Power Three

2

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

3

Probability

The Gambler

Kenny Rogers

20 Great Years

4

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

5

Outline continued

• conditional probability

• independence of two events

6

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

7

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

8

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

9

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

10

Uncertainty in Life

• Demography– Death rates – Marriage– divorce

11

Uncertainty in Life: US (CDC)

12

13

Probability of First Marriage by Age, Women: US (CDC)

14

Cohabitation: The Path to Marriage?: US(CDC)

15

Race/ethnicity Affects Duration of First Marriage

Properties of probabilities

• Nonnegative– example: p(h)

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

0

17

Flipping a coin twice: 4 elementary outcomes

heads

tails

heads

tails

heads

tails

h, h

h, t

t, h

t, t

18

Throwing Two Dice, 36 elementary outcomes

19

Larry Gonick and Woollcott Smith,The Cartoon Guideto Statistics

20

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

21

Event: white die equals one is the bottom row

Event: red die equals one is the right hand column

22

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

25

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

)11( RWp

27

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

29

1

2

3

4

5

6

1

2

3

4

5

6

Probability tree

2 rolls of a die:

36 elementary outcomes, of which 11 involve one or more sixes

30

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) ?

31

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

34

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

36

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

38

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

39

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

40

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

41

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.

42

Middle Aged men

Bald

P (Bald and MA) = 0.28

Not Bald

43

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

44

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

45

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)