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1 Econ 240A Power Three

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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

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3 Probability The Gambler Kenny Rogers 20 Great Years

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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

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5 Outline continued conditional probability independence of two events

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6 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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7 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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8 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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9 Uncertainty in Life Demography –Death rates –Marriage –divorce

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10 Uncertainty in Life: US (CDC)

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

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

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14 Race/ethnicity Affects Duration of First Marriage

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15 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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16 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 H H T Flip a coin

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17 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:

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18 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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19 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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Properties of probabilities Nonnegative –example: p(h) probabilities of elementary events sum to one –example p(h) + p(t) = 1

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21 Another Example: Toss Two Coins H1H1 T1T1 H2H2 T2T2 H, H H, T H2H2 T2T2 T, H T, T

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22 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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23 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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24 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:

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25 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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26 Throwing Two Dice, 36 elementary outcomes

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27 Larry Gonick and Woollcott Smith, The Cartoon Guide to Statistics

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28 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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29 Event: white die equals one is the bottom row Event: red die equals one is the right hand column

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30 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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31 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: Either the event A or the event B occurs or both do: The event A does not occur, i.e.not A:

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Probability statements Probability of either event A or event B –if the events are mutually exclusive, then probability of event B

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34 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

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36 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 –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

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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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39 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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40 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) ?

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Independence of two events p(A/B) = p(A) –i.e. if event A is not conditional on event B –then

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43 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 Distribution of a retail store purchases classified by amount and method of payment

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45 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

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47 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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48 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 = C. Determine the proportion of purchases made by credit card or debit card

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49 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) =

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50 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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51 Middle Aged men Bald P (Bald and MA) = 0.28 Not Bald

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52 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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53 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* *0.72 = =

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54 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)

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