INTRODUCTORY STATISTICS LECTURE 4 PROBABILITY

Transcription

1 INTRODUCTORY STATISTICS LECTURE 4 PROBABILITY

2 THE GREAT SCHLITZ CAMPAIGN 1981 Superbowl Broadcast of a live taste pitting Against key competitor: Michelob Subjects: 100 Michelob drinkers REF: SCHLITZBREWING.COM

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5 MICHELOB VS. SCHLITZ VS. REF: DAILYHIIT.COM.

6 MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability VS. REF: DAILYHIIT.COM.

7 MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability VS. probability of least say 40% of Michelob drinkers prefer Schlitz? REF: DAILYHIIT.COM.

8 MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability VS. probability of least say 40% of Michelob drinkers prefer Schlitz? 98% REF: DAILYHIIT.COM.

9 PROBABILITY GAMES REF:WIKIPEDIA.ORG

10 PROBABILITY GAMES Blackjack Yahtzee Backgammon Poker and many others REF:WIKIPEDIA.ORG

11 REF: NJAROUNDTHEWORLD.COM/

12 MIT blackjack team Students and professors from MIT Basic probability in 1980s, profit: 170$/hour Max in a single trip: 500K $ REF: NJAROUNDTHEWORLD.COM/

13 THE HISTORY Starts with Cardano Medical doctor Gambling addict Book on Games of Chance REF:WIKIPEDIA.ORG

14 FIRST PROBABILITY MODELS A coin is flipped What is the probability that a Tails show? We can say 0.5, if and only if the coin is unbiased What if outcomes are not equally likely?

15 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips b. exactly 3 heads in 100 coin flips c. exactly 3 heads in 1000 coin flips

16 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips b. exactly 3 heads in 100 coin flips c. exactly 3 heads in 1000 coin flips

17 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips b. exactly 3 heads in 100 coin flips c. exactly 3 heads in 1000 coin flips

18 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips b. exactly 3 heads in 100 coin flips c. exactly 3 heads in 1000 coin flips Total Outcomes: 2number of flips

19 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips C(10,3) / 2 10 b. exactly 3 heads in 100 coin flips C(100,3) / c. exactly 3 heads in 1000 coin flips C(1000,3) / a = 0.11, b = 9.46 x 10-29, c =1.11x

20 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips C(10,3) / 2 10 b. exactly 3 heads in 100 coin flips C(100,3) / c. exactly 3 heads in 1000 coin flips C(1000,3) / a = 0.11, b = 9.46 x 10-29, c =1.11x

21 REVIEW Which of the following events would you be most surprised by? a. exactly 3 heads in 10 coin flips C(10,3) / 2 10 b. exactly 3 heads in 100 coin flips C(100,3) / c. exactly 3 heads in 1000 coin flips C(1000,3) / a = 0.11, b = 9.46 x 10-29, c =1.11x

22 LAW OF LARGE NUMBERS Law of large numbers states that as more observations are collected, the proportion of occurrences with a particular outcome, A, converges to the probability of that outcome, P(A).

23 DISJOINT AND NON-DISJOINT OUTCOMES

24 DISJOINT AND NON-DISJOINT OUTCOMES Disjoint (mutually exclusive) outcomes: Cannot happen at the same time.

25 DISJOINT AND NON-DISJOINT OUTCOMES Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. The outcome of a single coin toss cannot be a head and a tail.

26 DISJOINT AND NON-DISJOINT OUTCOMES Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. The outcome of a single coin toss cannot be a head and a tail. A student both cannot fail and pass a class.

27 DISJOINT AND NON-DISJOINT OUTCOMES Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. The outcome of a single coin toss cannot be a head and a tail. A student both cannot fail and pass a class. Non-disjoint outcomes: Can happen at the same time.

28 DISJOINT AND NON-DISJOINT OUTCOMES Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. The outcome of a single coin toss cannot be a head and a tail. A student both cannot fail and pass a class. Non-disjoint outcomes: Can happen at the same time. A student can get an A in Stats and A in Econ in the same semester.

29 COMPLEMENTARY EVENTS

30 COMPLEMENTARY EVENTS Complementary events are two mutually exclusive events whose probabilities that add up to 1.

31 COMPLEMENTARY EVENTS Complementary events are two mutually exclusive events whose probabilities that add up to 1. A couple has one kid. If we know that the kid is not a boy, what is gender of this kid? {M,F} > Boy and girl are complementary outcomes.

32 COMPLEMENTARY EVENTS Complementary events are two mutually exclusive events whose probabilities that add up to 1. A couple has one kid. If we know that the kid is not a boy, what is gender of this kid? {M,F} > Boy and girl are complementary outcomes. A couple has two kids, if we know that they are not both girls, what are the possible gender combinations for these kids? {MM, MF, FM, FF}

33 INDEPENDENCE

34 INDEPENDENCE Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other.

35 INDEPENDENCE Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. Knowing that the coin landed on a head on the first toss does not provide any useful information for determining what the coin will land on in the second toss.

36 INDEPENDENCE Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. Knowing that the coin landed on a head on the first toss does not provide any useful information for determining what the coin will land on in the second toss. Outcomes of two tosses of a coin are independent.

37 INDEPENDENCE Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. Knowing that the coin landed on a head on the first toss does not provide any useful information for determining what the coin will land on in the second toss. Outcomes of two tosses of a coin are independent. Knowing that the first card drawn from a deck is an ace does provide useful information for determining the probability of drawing an ace in the second draw.

38 INDEPENDENCE Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. Knowing that the coin landed on a head on the first toss does not provide any useful information for determining what the coin will land on in the second toss. Outcomes of two tosses of a coin are independent. Knowing that the first card drawn from a deck is an ace does provide useful information for determining the probability of drawing an ace in the second draw. Outcomes of two draws from a deck of cards (without replacement) are dependent.

39 EXERCISE Between January 9-12, 2013, SurveyUSA interviewed a random sample of 500 NC residents asking them whether they think widespread gun ownership protects law abiding citizens from crime, or makes society more dangerous. 58% of all respondents said it protects citizens. 67% of White respondents, 28% of Black respondents, and 64% of Hispanic respondents shared this view. Which of the below is true? Opinion on gun ownership and race ethnicity are most likely (a) complementary (b) mutually exclusive (c) independent (d) dependent (e) disjoint

40 EXERCISE Between January 9-12, 2013, SurveyUSA interviewed a random sample of 500 NC residents asking them whether they think widespread gun ownership protects law abiding citizens from crime, or makes society more dangerous. 58% of all respondents said it protects citizens. 67% of White respondents, 28% of Black respondents, and 64% of Hispanic respondents shared this view. Which of the below is true? Opinion on gun ownership and race ethnicity are most likely (a) complementary (b) mutually exclusive (c) independent (d) dependent (e) disjoint

41 REVIEW

42 REVIEW

43 REVIEW Do the sum of probabilities of two disjoint events always add up to 1?

44 REVIEW Do the sum of probabilities of two disjoint events always add up to 1? Not necessarily, there may be more than 2 events in the sample space, e.g. party affiliation.

45 REVIEW Do the sum of probabilities of two disjoint events always add up to 1? Not necessarily, there may be more than 2 events in the sample space, e.g. party affiliation. Do the sum of probabilities of two complementary events always add up to 1?

46 REVIEW Do the sum of probabilities of two disjoint events always add up to 1? Not necessarily, there may be more than 2 events in the sample space, e.g. party affiliation. Do the sum of probabilities of two complementary events always add up to 1? Yes, that s the definition of complementary, e.g. heads and tails.

47 EXERCISE

48 EXERCISE If we were to randomly select 5 Texans, what is the probability that at least one is uninsured?

49 EXERCISE If we were to randomly select 5 Texans, what is the probability that at least one is uninsured? The sample space for the number of Texans who are uninsured would be: S = {0,1,2,3,4,5}

50 EXERCISE If we were to randomly select 5 Texans, what is the probability that at least one is uninsured? The sample space for the number of Texans who are uninsured would be: We are interested in instances where at least one person is uninsured: S = {0,1,2,3,4,5}

51 EXERCISE If we were to randomly select 5 Texans, what is the probability that at least one is uninsured? The sample space for the number of Texans who are uninsured would be: We are interested in instances where at least one person is uninsured: So we can divide up the sample space intro two categories: S = {0,at least one}

52 EXERCISE If we were to randomly select 5 Texans, what is the probability that at least one is uninsured? The sample space for the number of Texans who are uninsured would be: We are interested in instances where at least one person is uninsured: So we can divide up the sample space intro two categories:

53 EXERCISE (CONTINUED ) Since the probability of the sample space must add up to 1: Prob(at least 1 uninsured) = 1 Prob(none uninsured) = 1 - [ ( ) 5 ] = 0.77

54 EXERCISE Roughly 20% of undergraduates at a university are vegetarian or vegan. What is the probability that, among a random sample of 3 undergraduates, at least one is vegetarian or vegan? A B C D E

55 EXERCISE Roughly 20% of undergraduates at a university are vegetarian or vegan. What is the probability that, among a random sample of 3 undergraduates, at least one is vegetarian or vegan? A B C D E

56 CONDITIONAL PROBABILITY Researchers randomly assigned 72 chronic users of cocaine into three groups: desipramine (antidepressant), lithium (standard treatment for cocaine) and placebo. Results of the study are summarized below. REF: SRP/STATS/2_WAY_TBL_1.HTM

57 CONDITIONAL PROBABILITY Researchers randomly assigned 72 chronic users of cocaine into three groups: desipramine (antidepressant), lithium (standard treatment for cocaine) and placebo. Results of the study are summarized below. Relapse No Relapse Total Desipramin Lithium e Placebo Total REF: SRP/STATS/2_WAY_TBL_1.HTM

58 CONDITIONAL PROBABILITY What is the probability a patient relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total

59 CONDITIONAL PROBABILITY What is the probability a patient relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total

60 CONDITIONAL PROBABILITY What is the probability a patient relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total

61 CONDITIONAL PROBABILITY What is the probability a patient relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total P(relapsed) = 48/72 = 0.67

62 CONDITIONAL PROBABILITY What is the probability that a patient received desipramin and relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total

63 CONDITIONAL PROBABILITY What is the probability that a patient received desipramin and relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total P(relapsed) = 10/72 = 0.14

64 CONDITIONAL PROBABILITY What is the probability that a patient received desipramin and relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total P(relapsed) = 10/72 = 0.14

65 CONDITIONAL PROBABILITY What is the probability that a patient received desipramin and relapsed? Relapse No Relapse Total Desipramin Lithium e Placebo Total P(relapsed) = 10/72 = 0.14

66 CONDITIONAL PROBABILITY Conditional probability The conditional probability of the outcome of interest A, given condition B, is calculated as P (A B) = P (A and B) P (B)

67 Probability that patients took desipramin relapsed P( relapsed disepramin) = Relapse No Relapse Total Desipramin Lithium e Placebo Total

68 Probability that patients took desipramin relapsed P( relapsed disepramin) = P (relapse and disepramin) P (disepramin) Relapse No Relapse Total Desipramin Lithium e Placebo Total

69 Probability that patients took desipramin relapsed P( relapsed disepramin) = P (relapse and disepramin) P (disepramin) Relapse No Relapse Total Desipramin Lithium e Placebo Total

70 Probability that patients took desipramin relapsed P( relapsed disepramin) = P (relapse and disepramin) P (disepramin) Relapse No Relapse Total Desipramin Lithium e Placebo Total

71 Probability that patients took desipramin relapsed P( relapsed disepramin) = P (relapse and disepramin) P (disepramin) Relapse No Relapse Total Desipramin Lithium e Placebo Total /24 = 0.42

72 INDEPENDENT EVENTS Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. If A and B are independent events, P(A B) = P(A) If P(A B) = P(A), then A and B are independent events

73 BREAST CANCER American Cancer Society estimates that about 1.7% of women have breast cancer. CANCER/ CANCERBASICS/ CANCER-PREVALENCE Susan G. Komen For The Cure Foundation states that mammography correctly identifies about 78% of women who truly have breast cancer. WW5.KOMEN.ORG/ BREASTCANCER/ ACCURACYOFMAMMOGRAMS.HTML An article published in 2003 suggests that up to 10% of all mammograms result in false positives for patients who do not have cancer. PMC/ ARTICLES/ PMC136094

74 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, positive, 0.78 negative, *0.78 = *0.22 = no cancer, positive, 0.1 negative, *0.1 = *0.9 =

75 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, positive, 0.78 negative, *0.78 = *0.22 = no cancer, positive, 0.1 negative, *0.1 = *0.9 =

76 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, positive, 0.78 negative, *0.78 = *0.22 = no cancer, positive, 0.1 negative, *0.1 = *0.9 =

77 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, positive, 0.78 negative, *0.78 = *0.22 = no cancer, positive, 0.1 negative, *0.1 = *0.9 =

78 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, positive, 0.78 negative, *0.78 = *0.22 = no cancer, positive, 0.1 negative, *0.1 = *0.9 =

79 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, positive, 0.78 negative, *0.78 = *0.22 = no cancer, positive, 0.1 negative, *0.1 = *0.9 =

80 BREAST CANCER When a patient goes through breast cancer screening there are two competing claims: patient had cancer and patient doesn t have cancer. If a mammogram yields a positive result, what is the probability that patient actually has cancer? Cancer status Test result cancer, no cancer, positive, 0.78 negative, 0.22 positive, *0.78 = *0.22 = *0.1 = P(C +) P(C and +) = P(+) = = 0.12 negative, *0.9 =

81 BAYES THEOREM Bayes Theorem: P(outcome A of variable 1 outcome B of variable 2) = P(B A 1 )P(A 1 ) P(B A 1 )P(A 1 ) + P(B A 2 )P(A 2 ) + + P(B A k )P(A k ) where A1, A2,, Ak represent all other possible outcomes of variable 1.

82 EXAMPLE

83 EXAMPLE A common epidemiological model for the spread of diseases is the SIR model, where the population is partitioned into three groups: Susceptible, Infected, and Recovered. This is a reasonable model for diseases like chickenpox where a single infection usually provides immunity to subsequent infections. Sometimes these diseases can also be difficult to detect.

84 EXAMPLE A common epidemiological model for the spread of diseases is the SIR model, where the population is partitioned into three groups: Susceptible, Infected, and Recovered. This is a reasonable model for diseases like chickenpox where a single infection usually provides immunity to subsequent infections. Sometimes these diseases can also be difficult to detect. Imagine a population in the midst of an epidemic where 60% of the population is considered susceptible, 10% is infected, and 30% is recovered. The only test for the disease is accurate 95% of the time for susceptible individuals, 99% for infected individuals, but 65% for recovered individuals. (Note: In this case accurate means returning a negative result for susceptible and recovered individuals and a positive result for infected individuals).

85 EXAMPLE A common epidemiological model for the spread of diseases is the SIR model, where the population is partitioned into three groups: Susceptible, Infected, and Recovered. This is a reasonable model for diseases like chickenpox where a single infection usually provides immunity to subsequent infections. Sometimes these diseases can also be difficult to detect. Imagine a population in the midst of an epidemic where is considered susceptible, for the disease is accurate infected individuals accurate means returning a negative result for susceptible and recovered individuals and a positive result for infected individuals). Draw a probability tree to reflect the information given above. If the individual has tested positive, what is the probability that they are actually infected?

86 ` 60%: susceptible 10%: infected 30%:recovered. TEST 95% : susceptible 99% : infected 65% : recovered Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

87 ` 60%: susceptible 10%: infected 30%:recovered. TEST 95% : susceptible 99% : infected 65% : recovered Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

88 ` 60%: susceptible 10%: infected 30%:recovered. TEST 95% : susceptible 99% : infected 65% : recovered Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

89 ` 60%: susceptible 10%: infected 30%:recovered. TEST 95% : susceptible 99% : infected 65% : recovered Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

90 EXERCISE (CONT.) Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

91 EXERCISE (CONT.) Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

92 EXERCISE (CONT.) P(inf +) = P(inf and +) P(+) = Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

93 EXERCISE (CONT.) P(inf +) = P(inf and +) P(+) = Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

94 EXERCISE (CONT.) P(inf +) = P(inf and +) P(+) = Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

95 EXERCISE (CONT.) P(inf +) = P(inf and +) P(+) = Group Test result susceptible, 0.6 infected, 0.1 recovered, 0.3 positive, 0.05 negative, 0.95 positive, 0.99 negative, 0.01 positive, 0.35 negative,

96 RANDOM VARIABLES

97 RANDOM VARIABLES A random variable is a numeric quantity whose value depends on the outcome of a random event

98 RANDOM VARIABLES A random variable is a numeric quantity whose value depends on the outcome of a random event We use a capital letter, like X, to denote a random variable.

99 RANDOM VARIABLES A random variable is a numeric quantity whose value depends on the outcome of a random event We use a capital letter, like X, to denote a random variable. The values of a random variable are denoted with a lower case letter, in this case x For example, P( X=x)

100 EXPECTATION We are often interested in the average outcome of a random variable. We call this the expected value (mean), and it is a weighted average of the possible outcomes µ = E(X) = kx xp (X = x i ) i=1

101 A DICE GAME

102 A DICE GAME Enter to game: 3$

103 A DICE GAME Enter to game: 3$ You roll a fair dice, whatever value you roll, you get its dollar equivalent

104 A DICE GAME Enter to game: 3$ You roll a fair dice, whatever value you roll, you get its dollar equivalent E.g: If you roll a 2, you get 2$, if you roll 5 you get 5$

105 A DICE GAME Enter to game: 3$ You roll a fair dice, whatever value you roll, you get its dollar equivalent E.g: If you roll a 2, you get 2$, if you roll 5 you get 5$ If you play this game 3000 times, how much money would you expect to lose or win?

106 A DICE GAME Outcomes = {1,2,3,4,5,6} After 1000 times, you ll get almost equal amounts of 1 s, 2 s, 6 s, that is 3000/6 = 500 times Total Earning = (1 x x x 500)= 10,500 Total Cost = 3 x 3000 = 9,000 Expected Profit = 10,500-9,000 = 1,500 $

107 EXPECTATION

108 EXPECTATION

109 EXPECTATION Dice X P(X) X P(X)

110 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6

111 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6 2 $2 1/6 2/6

112 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6 2 $2 1/6 2/6 3 $3 1/6 3/6

113 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6 2 $2 1/6 2/6 3 $3 1/6 3/6 4 $4 1/6 4/6

114 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6 2 $2 1/6 2/6 3 $3 1/6 3/6 4 $4 1/6 4/6 5 $5 1/6 5/6

115 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6 2 $2 1/6 2/6 3 $3 1/6 3/6 4 $4 1/6 4/6 5 $5 1/6 5/6 6 $6 1/6 6/6

116 EXPECTATION Dice X P(X) X P(X) 1 $1 1/6 1/6 2 $2 1/6 2/6 3 $3 1/6 3/6 4 $4 1/6 4/6 5 $5 1/6 5/6 6 $6 1/6 6/6 Total E(X) = 21/6 = 3.5

117 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE In a game of cards you win $1 if you draw a heart, $5 if you draw an ace (including the ace of hearts), $10 if you draw the king of spades and nothing for any other card you draw. Write the probability model for your winnings, and calculate your expected winning.

118 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE $1, a heart- $5, an ace $10, king of spades - $0, any other card Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

119 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE $1, a heart- $5, an ace $10, king of spades - $0, any other card Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

120 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

121 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE Probability X Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

122 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE Probability X Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

123 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE Probability X Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

124 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE Probability X Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

125 EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE Probability X Event X P(X) XP(X) Heart (not ace) Ace King of spades All else Total E(X) =

126 VARIANCE OF A RANDOM VARIABLE

127 VARIANCE OF A RANDOM VARIABLE The variable in winnings

128 VARIANCE OF A RANDOM VARIABLE The variable in winnings

129 VARIANCE OF A RANDOM VARIABLE The variable in winnings

130 VARIANCE OF A RANDOM VARIABLE The variable in winnings

131 VARIANCE OF A RANDOM VARIABLE The variable in winnings

132 LINEAR COMBINATIONS a linear combination of random variables X and Y is ax + by such as, 3X + 5Y 0.23 X - 32Y What is E(aX + by)=?

133 LINEAR COMBINATIONS a linear combination of random variables X and Y is ax + by such as, 3X + 5Y 0.23 X - 32Y What is E(aX + by)=? E(aX + by )=ae(x)+be(y )

134 LINEAR COMBINATIONS The variability of linear combinations of two independent random variables

135 LINEAR COMBINATIONS The variability of linear combinations of two independent random variables Var(aX + by )=a 2 Var(X)+b 2 Var(Y )

136 EXERCISE A casino game costs $5 to play. If you draw first a red card, then you get to draw a second card. If the second card is the ace of hearts, you win $500. If not, you don t win anything, i.e. lose your $5. What is your expected profits/losses from playing this game?

137 EXERCISE A casino game costs $5 to play. If you draw first a red card, then you get to draw a second card. If the second card is the ace of hearts, you win $500. If not, you don t win anything, i.e. lose your $5. What is your expected profits/losses from playing this game?

138 EXERCISE A casino game costs $5 to play. If you draw first a red card, then you get to draw a second card. If the second card is the ace of hearts, you win $500. If not, you don t win anything, i.e. lose your $5. What is your expected profits/losses from playing this game?

139 EXERCISE A casino game costs $5 to play. If you draw first a red card, then you get to draw a second card. If the second card is the ace of hearts, you win $500. If not, you don t win anything, i.e. lose your $5. What is your expected profits/losses from playing this game?

140 EXERCISE A casino game costs $5 to play. If you draw first a red card, then you get to draw a second card. If the second card is the ace of hearts, you win $500. If not, you don t win anything, i.e. lose your $5. What is your expected profits/losses from playing this game?

141 SIMPLIFYING RANDOM VARIABLES

142 OVERVIEW Probability Random Variables Continuous Distributions MOST OF THE SLIDES ADOPTED FROM OPENINTRO STATS BOOK.

143 CONTINOUS DISTRIBUTIONS Below is a histogram of the distribution of heights of US adults. The proportion of data that falls in the shaded bins gives the probability that a randomly sampled US adult is between 180 cm and 185 cm (about 5 11 to 6 1 ).

144 FROM HISTOGRAMS TO CONTINUOUS DISTRIBUTIONS Since height is a continuous numerical variable, its probability density function is a smooth curve.

145 FROM HISTOGRAMS TO CONTINUOUS DISTRIBUTIONS Therefore, the probability that a randomly sampled US adult is between 180 cm and 185 cm can also be estimated as the shaded area under the curve.

146 FROM HISTOGRAMS TO CONTINUOUS DISTRIBUTIONS Therefore, the probability that a randomly sampled US adult is between 180 cm and 185 cm can also be estimated as the shaded area under the curve.

147 FROM HISTOGRAMS TO CONTINUOUS DISTRIBUTIONS Since continuous probabilities are estimated as the area under the curve, the probability of a person being exactly 180 cm (or any exact value) is defined as 0.