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Introduction
• Nothing in life is certain• We gauge the chances of
successful outcomes in business, medicine, weather, and other everyday situations such as the lottery (recall the birthday problem)
History
• For most of human history, probability, the formal study of the laws of chance, has been used for only one thing: gambling
History (cont.)• Nobody knows exactly when
gambling began; goes back at least as far as ancient Egypt where 4-sided “astragali” (made from animal heelbones) were used
History (cont.)• The Roman emperor Claudius
(10BC-54AD) wrote the first known treatise on gambling.
• The book “How to Win at Gambling” was lost.
Rule 1: Let Caesar win IVout of V times
Approaches to Probability
• Relative frequencyevent probability = x/n, where x=# of occurrences of event of interest, n=total # of observations
• Coin, die tossing; nuclear power plants?
• Limitationsrepeated observations not practical
Approaches to Probability (cont.)
• Subjective probabilityindividual assigns prob. based on personal experience, anecdotal evidence, etc.
• Classical approachevery possible outcome has equal probability (more later)
Basic Definitions
• Experiment: act or process that leads to a single outcome that cannot be predicted with certainty
• Examples:1. Toss a coin2. Draw 1 card from a standard deck of
cards3. Arrival time of flight from Atlanta to
RDU
Basic Definitions (cont.)
• Sample space: all possible outcomes of an experiment. Denoted by S
• Event: any subset of the sample space S;typically denoted A, B, C, etc.Simple event: event with only 1 outcomeNull event: the empty set Certain event: S
Examples
1. Toss a coin onceS = {H, T}; A = {H}, B = {T} simple events
2. Toss a die once; count dots on upper faceS = {1, 2, 3, 4, 5, 6}A=even # of dots on upper face={2, 4, 6}B=3 or fewer dots on upper face={1, 2, 3}
Laws of Probability (cont.)
3. P(A’ ) = 1 - P(A)For an event A, A’ is the complement of A; A’ is everything in S that is not in A.
AA'
S
Birthday Problem• What is the smallest number of
people you need in a group so that the probability of 2 or more people having the same birthday is greater than 1/2?
• Answer: 23No. of people 23 30 40 60Probability .507.706.891.994
Example: Birthday Problem
• A={at least 2 people in the group have a common birthday}
• A’ = {no one has common birthday}
502.498.1)'(1)(
498.365
343
365
363
365
364)'(
:23365
363
365
364)'(:3
APAPso
AP
people
APpeople
Laws of Probability (cont.)
Addition Rule for Disjoint Events:
4. If A and B are disjoint events, then
P(A B) = P(A) + P(B)
Laws of Probability (cont.)
General Addition Rule
6. For any two events A and B
P(A B) = P(A) + P(B) – P(A B)
Example: toss a fair die once
• S = {1, 2, 3, 4, 5, 6}• A = even # appears = {2, 4, 6}• B = 3 or fewer = {1, 2, 3}• P(A B) = P(A) + P(B) - P(A B)
=P({2, 4, 6}) + P({1, 2, 3}) - P({2})
= 3/6 + 3/6 - 1/6 = 5/6
Laws of Probability: Summary
• 1. 0 P(A) 1 for any event A• 2. P() = 0, P(S) = 1• 3. P(A’) = 1 – P(A)• 4. If A and B are disjoint events, then
P(A B) = P(A) + P(B)• 5. If A and B are independent events,
thenP(A B) = P(A) × P(B)
• 6. For any two events A and B,P(A B) = P(A) + P(B) – P(A B)
Assigning Probabilities
• If an experiment has N outcomes, then each outcome has probability 1/N of occurring
• If an event A1 has n1 outcomes, then
P(A1) = n1/N
Product Rule for Ordered Pairs
• A student wishes to commute to a junior college for 2 years and then commute to a state college for 2 years. Within commuting distance there are 4 junior colleges and 3 state colleges. How many junior college-state college pairs are available to her?
Product Rule for Ordered Pairs
• junior colleges: 1, 2, 3, 4• state colleges a, b, c• possible pairs:(1, a) (1, b) (1, c)(2, a) (2, b) (2, c)(3, a) (3, b) (3, c)(4, a) (4, b) (4, c)
Product Rule for Ordered Pairs
• junior colleges: 1, 2, 3, 4• state colleges a, b, c• possible pairs:(1, a) (1, b) (1, c)(2, a) (2, b) (2, c)(3, a) (3, b) (3, c)(4, a) (4, b) (4, c)
4 junior colleges3 state collegestotal number of possiblepairs = 4 x 3 = 12
4 junior colleges3 state collegestotal number of possiblepairs = 4 x 3 = 12
Product Rule for Ordered Pairs
• junior colleges: 1, 2, 3, 4• state colleges a, b, c• possible pairs:(1, a) (1, b) (1, c) (2, a) (2, b) (2, c)(3, a) (3, b) (3, c)(4, a) (4, b) (4, c)
In general, if there are n1 waysto choose the first element ofthe pair, and n2 ways to choosethe second element, then the number of possible pairs isn1n2. Here n1 = 4, n2 = 3.
In general, if there are n1 waysto choose the first element ofthe pair, and n2 ways to choosethe second element, then the number of possible pairs isn1n2. Here n1 = 4, n2 = 3.
Counting in “Either-Or” Situations• NCAA Basketball Tournament: how
many ways can the “bracket” be filled out?
1. How many games?2. 2 choices for each game3. Number of ways to fill out the bracket:
263 = 9.2 × 1018
• Earth pop. about 6 billion; everyone fills out 1 million different brackets
• Chances of getting all games correct is about 1 in 1,000
Counting Example
• Pollsters minimize lead-in effect by rearranging the order of the questions on a survey
• If Gallup has a 5-question survey, how many different versions of the survey are required if all possible arrangements of the questions are included?
Solution• There are 5 possible choices for the
first question, 4 remaining questions for the second question, 3 choices for the third question, 2 choices for the fourth question, and 1 choice for the fifth question.
• The number of possible arrangements is therefore
5 4 3 2 1 = 120
Efficient Methods for Counting Outcomes
• Factorial Notation:n!=12 … n
• Examples1!=1; 2!=12=2; 3!= 123=6; 4!
=24;5!=120;• Special definition: 0!=1
Factorials with calculators and Excel
• Calculator: non-graphing: x ! (second function)graphing: bottom p. 9 T I Calculator Commands(math button)
• Excel:Paste: math, fact
Factorial Examples• 20! = 2.43 x 1018
• 1,000,000 seconds?• About 11.5 days• 1,000,000,000 seconds?• About 31 years• 31 years = 109 seconds• 1018 = 109 x 109
• 31 x 109 years = 109 x 109 = 1018 seconds
• 20! is roughly the age of the universe in seconds
Permutations
A B C D E• How many ways can we choose 2
letters from the above 5, without replacement, when the order in which we choose the letters is important?
• 5 4 = 20
Permutations with calculator and Excel
• Calculatornon-graphing: nPr
• Graphingp. 9 of T I Calculator Commands(math button)
• ExcelPaste: Statistical, Permut
Combinations
A B C D E• How many ways can we choose 2
letters from the above 5, without replacement, when the order in which we choose the letters is not important?
• 5 4 = 20 when order important• Divide by 2: (5 4)/2 = 10 ways
ST 101 Powerball Lottery
From the numbers 1 through 20,choose 6 different numbers.
Write them on a piece of paper.
Chances of Winning?
760,38!6)!620(
!20
ies?possibilit ofNumber
important.not order t,replacemen
without 20, from numbers 6 Choose
620206
C
North Carolina Powerball Lottery
Prior to Jan. 1, 2009 After Jan. 1, 2009
:
55!3,478,761
5!50!
:
42!42
1!41!
3,478,761*42
146,107,962
5 from 1- 55
1 from 1- 42 (p'ball #)
:
59!5,006,386
5!54!
:
39!39
1!38!
5,006,386*39
195,249,054
5 from 1- 59
1 from 1- 39 (p'ball #)
Visualize Your Lottery Chances
• How large is 195,249,054?• $1 bill and $100 bill both 6” in length
• 10,560 bills = 1 mile• Let’s start with 195,249,053 $1 bills
and one $100 bill …• … and take a long walk, putting
down bills end-to-end as we go
Chances of Winning NC Powerball Lottery?
• Remember: one of the bills you put down is a $100 bill; all others are $1 bills
• Your chance of winning the lottery is the same as bending over and picking up the $100 bill while walking the route blindfolded.
Example: Illinois State Lottery
balls) pong pingmillion 16.5 house, ft (1200
months) 10in second 1about (
165,827,25!6!48
!54
importantnot order t;replacemen
withoutnumbers 54 from numbers 6 Choose
2
654 C
Example: AIDS Testing• V={person has HIV}; CDC: P(V)=.006• +: test outcome is positive (test
indicates HIV present)• -: test outcome is negative• clinical reliabilities for a new HIV test:
1. If a person has the virus, the test result will be positive with probability .999
2. If a person does not have the virus, the test result will be negative with probability .990
Probability Tree Approach
• A probability tree is a useful way to visualize this problem and to find the desired probability.
Question 1 Answer
• What is the probability that a randomly selected person will test positive?
• P(+) = .00599 + .00994 = .01593
Question 2
• If your test comes back positive, what is the probability that you have HIV?(Remember: we know that if a person has the virus, the test result will be positive with probability .999; if a person does not have the virus, the test result will be negative with probability .990).
• Looks very reliable
Question 2 Answer
Answertwo sequences of branches lead to positive test; only 1 sequence represented people who have HIV.
P(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376
Summary• Question 1:• P(+) = .00599 + .00994 = .01593• Question 2: two sequences of
branches lead to positive test; only 1 sequence represented people who have HIV.
P(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376
Recap• We have a test with very high clinical
reliabilities:1. If a person has the virus, the test result will be
positive with probability .9992. If a person does not have the virus, the test
result will be negative with probability .990
• But we have extremely poor performance when the test is positive:
P(person has HIV given that test is positive) =.376
• In other words, 62.4% of the positives are false positives! Why?
• When the characteristic the test is looking for is rare, most positives will be false.