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Transcript of June 3, 2008Stat 111 - Lecture 6 - Probability1 Probability Introduction to Probability, Conditional...
June 3, 2008 Stat 111 - Lecture 6 - Probability 1
Probability
Introduction to Probability,
Conditional Probability and
Random Variables
Statistics 111 - Lecture 6
June 3, 2008 Stat 111 - Lecture 6 - Probability 2
Administrative Note
• Homework 2 due Monday, June 8th
– Look at the questions now!
• Prepare to have your minds blown today
June 2, 2008 Stat 111 - Lecture 6 - Introduction 3
Course Overview
Collecting Data
Exploring DataProbability Intro.
Inference
Comparing Variables Relationships between Variables
Means Proportions Regression Contingency Tables
June 3, 2008 Stat 111 - Lecture 6 - Probability 4
Why do we need Probability?
• We have several graphical and numerical statistics for summarizing our data
• We want to make probability statements about the significance of our statistics
• Eg. In Stat111, mean(height) = 66.7 inches • What is the chance that the true height of Penn
students is between 60 and 70 inches?
• Eg. r = -0.22 for draft order and birthday• What is the chance that the true correlation is
significantly different from zero?
June 3, 2008 Stat 111 - Lecture 6 - Probability 5
Deterministic vs. Random Processes
• In deterministic processes, the outcome can be predicted exactly in advance• Eg. Force = mass x acceleration. If we are given
values for mass and acceleration, we exactly know the value of force
• In random processes, the outcome is not known exactly, but we can still describe the probability distribution of possible outcomes • Eg. 10 coin tosses: we don’t know exactly how
many heads we will get, but we can calculate the probability of getting a certain number of heads
June 3, 2008 Stat 111 - Lecture 6 - Probability 6
Events
• An event is an outcome or a set of outcomes of a random process
Example: Tossing a coin three times
Event A = getting exactly two heads = {HTH, HHT, THH}
Example: Picking real number X between 1 and 20
Event A = chosen number is at most 8.23 = {X ≤ 8.23}
Example: Tossing a fair dice
Event A = result is an even number = {2, 4, 6}
• Notation: P(A) = Probability of event A• Probability Rule 1:
0 ≤ P(A) ≤ 1 for any event A
June 3, 2008 Stat 111 - Lecture 6 - Probability 7
Sample Space
• The sample space S of a random process is the set of all possible outcomes Example: one coin toss
S = {H,T} Example: three coin tosses
S = {HHH, HTH, HHT, TTT, HTT, THT, TTH, THH}Example: roll a six-sided dice
S = {1, 2, 3, 4, 5, 6}Example: Pick a real number X between 1 and 20
S = all real numbers between 1 and 20
• Probability Rule 2: The probability of the whole sample space is 1
P(S) = 1
June 3, 2008 Stat 111 - Lecture 6 - Probability 8
Combinations of Events• The complement Ac of an event A is the event that A
does not occur• Probability Rule 3:
P(Ac) = 1 - P(A)• The union of two events A and B is the event that
either A or B or both occurs• The intersection of two events A and B is the event
that both A and B occur
Event A Complement of A Union of A and B Intersection of A and B
June 3, 2008 Stat 111 - Lecture 6 - Probability 9
Disjoint Events• Two events are called disjoint if they can not
happen at the same time • Events A and B are disjoint means that the
intersection of A and B is zero
• Example: coin is tossed twice • S = {HH,TH,HT,TT}• Events A={HH} and B={TT} are disjoint • Events A={HH,HT} and B = {HH} are not disjoint
• Probability Rule 4: If A and B are disjoint events then
P(A or B) = P(A) + P(B)
June 3, 2008 Stat 111 - Lecture 6 - Probability 10
Independent events• Events A and B are independent if knowing that A
occurs does not affect the probability that B occurs
• Example: tossing two coinsEvent A = first coin is a head
Event B = second coin is a head
• Disjoint events cannot be independent!• If A and B can not occur together (disjoint), then knowing that
A occurs does change probability that B occurs
• Probability Rule 5: If A and B are independent
P(A and B) = P(A) x P(B)
Independent
multiplication rule for independent events
June 3, 2008 Stat 111 - Lecture 6 - Probability 11
Equally Likely Outcomes Rule
• If all possible outcomes from a random process have the same probability, then
• P(A) = (# of outcomes in A)/(# of outcomes in S)
• Example: One Dice Tossed
P(even number) = |2,4,6| / |1,2,3,4,5,6|
• Note: equal outcomes rule only works if the number of outcomes is “countable”• Eg. of an uncountable process is sampling any fraction between
0 and 1. Impossible to count all possible fractions !
June 3, 2008 Stat 111 - Lecture 6 - Probability 12
Combining Probability Rules Together
• Initial screening for HIV in the blood first uses an enzyme immunoassay test (EIA)
• Even if an individual is HIV-negative, EIA has probability of 0.006 of giving a positive result
• Suppose 100 people are tested who are all HIV-negative. What is probability that at least one will show positive on the test?
• First, use complement rule:P(at least one positive) = 1 - P(all negative)
June 3, 2008 Stat 111 - Lecture 6 - Probability 13
• Now, we assume that each individual is independent and use the multiplication rule for independent events:
P(all negative) = P(test 1 negative) ×…× P(test 100 negative)
• P(test negative) = 1 - P(test positive) = 0.994
P(all negative) = 0.994 ×…× 0.994 = (0.994)100
• So, we finally we have
P(at least one positive) =1− (0.994)100 = 0.452
Combining Probability Rules Together
June 3, 2008 Stat 111 - Lecture 6 - Probability 14
Curse of the Bambino:
Boston Red Sox traded Babe Ruth after 1918 and did not win a World Series again until 2004 (86 years later)
• What are the chances that a team will go 86 years without winning a world series?
• Simplifying assumptions: • Baseball has always had 30 teams • Each team has equal chance of winning each year
June 3, 2008 Stat 111 - Lecture 6 - Probability 15
Curse of the Bambino
• With 30 teams that are “equally likely” to win in a year, we have
P(no WS in a year) = 29/30 = 0.97
• If we also assume that each year is independent, we can use multiplication rule
P(no WS in 86 years)
= P(no WS in year 1) x… xP(no WS in year 86)
= (0.97) x… x (0.97)
= (0.97)86 = 0.05 (only 5% chance!)
Break
June 3, 2008 Stat 111 - Lecture 6 - Probability 16
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
17
Outline
• Moore, McCabe and Craig: Section 4.3,4.5
• Conditional Probability
• Discrete Random Variables
• Continuous Random Variables
• Properties of Random Variables• Means of Random Variables• Variances of Random Variables
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Conditional Probabilities
• The notion of conditional probability can be found in many different types of problems
• Eg. imperfect diagnostic test for a disease
• What is probability that a person has the disease? Answer: 40/100 = 0.4
• What is the probability that a person has the disease given that they tested positive?More Complicated !
Disease + Disease - Total
Test + 30 10 40
Test - 10 50 60
Total 40 60 100
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Definition: Conditional Probability
• Let A and B be two events in sample space
• The conditional probability that event B occurs given that event A has occurred is:
P(A|B) = P(A and B) / P(B)
• Eg. probability of disease given test positive
P(disease +| test +) = P(disease + and test +) / P(test +) = (30/100)/(40/100) =.75
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Independent vs. Non-independent Events
• If A and B are independent, then
P(A and B) = P(A) x P(B)
which means that conditional probability is:
P(B | A) = P(A and B) / P(A) = P(A)P(B)/P(A) = P(B)
• We have a more general multiplication rule for events that are not independent:
P(A and B) = P(B | A) × P(A)
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Random variables
• A random variable is a numerical outcome of a random process or random event
• Example: three tosses of a coin • S = {HHH,THH,HTH,HHT,HTT,THT,TTH,TTT}• Random variable X = number of observed tails• Possible values for X = {0,1, 2, 3}
• Why do we need random variables?• We use them as a model for our observed data
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Discrete Random Variables
• A discrete random variable has a finite or countable number of distinct values
• Discrete random variables can be summarized by listing all values along with the probabilities• Called a probability distribution
• Example: number of members in US families
X 2 3 4 5 6 7
P(X) 0.413 0.236 0.211 0.090 0.032 0.018
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Another Example
• Random variable X = the sum of two dice• X takes on values from 2 to 12
• Use “equally-likely outcomes” rule to calculate the probability distribution:
• If discrete r.v. takes on many values, it is better to use a probability histogram
X 2 3 4 5 6 7 8 9 10 11 12
# of
Outcomes
1 2 3 4 5 6 5 4 3 2 1
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Probability Histograms
• Probability histogram of sum of two dice:
• Using the disjoint addition rule, probabilities for discrete random variables are calculated by adding up the “bars” of this histogram:
P(sum > 10) = P(sum = 11) + P(sum = 12) = 3/36
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Continuous Random Variables
• Continuous random variables have a non-countable number of values
• Can’t list the entire probability distribution, so we use a density curve instead of a histogram
• Eg. Normal density curve:
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Calculating Continuous Probabilities
• Discrete case: add up bars from probability histogram• Continuous case: we have to use integration to
calculate the area under the density curve:
• Although it seems more complicated, it is often easier to integrate than add up discrete “bars”• If a discrete r.v. has many possible values, we often
treat that variable as continuous instead
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Example: Normal Distribution
We will use the normal distribution throughout
this course for two reasons:1. It is usually good approximation to real data
2. We have tables of calculated areas under the normal curve, so we avoid doing integration!
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Mean of a Random Variable• Average of all possible values of a random
variable (often called expected value)• Notation: don’t want to confuse random
variables with our collected data variables
= mean of random variable x = mean of a data variable
• For continuous r.v, we again need integration to calculate the mean
• For discrete r.v., we can calculate the mean by hand since we can list all probabilities
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Mean of Discrete random variables
• Mean is the sum of all possible values, with each value weighted by its probability:
μ = Σ xi*P(xi) = x1*P(x1) + … + x12*P(x12)
• Example: X = sum of two dice
μ = 2 (1/36) + 3 (2/36) + 4 (3/36) +⋅ ⋅ ⋅ …+12 (1/36)⋅ = 252/36 = 7
X 2 3 4 5 6 7 8 9 10 11 12
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Variance of a Random Variable• Spread of all possible values of a random
variable around its mean• Again, we don’t want to confuse random
variables with our collected data variables:
2 = variance of random variables2 = variance of a data variable
• For continuous r.v, again need integration to calculate the variance
• For discrete r.v., can calculate the variance by hand since we can list all probabilities
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Variance of Discrete r.v.s
• Variance is the sum of the squared deviations away from the mean of all possible values, weighted by the values probability:
μ = Σ(xi-μ)*P(xi) = (x1-μ)*P(x1) + … + (x12-μ)*P(x12)
• Example: X = sum of two dice
σ2 = (2 - 7)2 (1/36) +⋅ (3− 7)2 (2/36) +⋅ …+(12 - 7)2 (1/36)⋅ = 210/36 = 5.83
X 2 3 4 5 6 7 8 9 10 11 12
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
June 4, 2008 Stat 111 - Lecture 6 - Random Variables
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Next Class - Lecture 7
• Standardization and the Normal Distribution
• Moore and McCabe: Section 4.3,1.3