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Probability. Formal study of uncertainty The engine that drives statistics. 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.

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Probability
Probability

Formal study of uncertainty

The engine that drives statistics


Introduction
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
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
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 cont1

Rule 1: Let Caesar win IV

out of V times

History (cont.)

  • The Roman emperor Claudius (10BC-54AD) wrote the first known treatise on gambling.

  • The book “How to Win at Gambling” was lost.


Approaches to probability
Approaches to Probability

  • Relative frequency

    event probability = x/n,

    where x=# of occurrences of event of interest, n=total # of observations

  • Coin, die tossing; nuclear power plants?

  • Limitations

    repeated observations not practical


Approaches to probability cont
Approaches to Probability (cont.)

  • Subjective probability

    individual assigns prob. based on personal experience, anecdotal evidence, etc.

  • Classical approach

    every possible outcome has equal probability (more later)


Basic definitions
Basic Definitions

  • Experiment: act or process that leads to a single outcome that cannot be predicted with certainty

  • Examples:

    1.Toss a coin

    2.Draw 1 card from a standard deck of cards

    3.Arrival time of flight from Atlanta to RDU


Basic definitions cont
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 outcome

    Null event: the empty set F

    Certain event: S


Examples
Examples

1.Toss a coin once

S = {H, T}; A = {H}, B = {T} simple events

2.Toss a die once; count dots on upper face

S = {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
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.

S

A'

A


Birthday problem
Birthday Problem

  • What is the smallest number of people you need in a group so that the probability of 2 or morepeople having the same birthday is greater than 1/2?

  • Answer: 23

    No. of people23304060

    Probability.507.706.891.994


Example birthday problem
Example: Birthday Problem

  • A={at least 2 people in the group have a common birthday}

  • A’ = {no one has common birthday}


Unions and intersections
Unions and Intersections

S

AÇB

A

B

AÈB


Mutually exclusive events
Mutually Exclusive Events

  • Mutually exclusive events-no outcomes from S in common

A Ç B = Æ

S

A

B


Laws of probability cont1
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)


Probability


Laws of probability cont2
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)


P a b p a p b p a b
P(AÈB)=P(A) + P(B) - P(A Ç B)

S

AÇB

A

B


Example toss a fair die once
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 È B) = P(A) + P(B) - P(A Ç B)

    =P({2, 4, 6}) + P({1, 2, 3}) - P({2})

    = 3/6 + 3/6 - 1/6 = 5/6


Laws of probability summary
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, then

    P(A  B) = P(A) × P(B)

  • 6. For any two events A and B,

    P(A  B) = P(A) + P(B) – P(A  B)


Probability models
Probability Models

The Equally Likely Approach

(also called the Classical Approach)


Assigning probabilities
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
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 pairs1
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 pairs2
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 colleges

3 state colleges

total number of possible

pairs = 4 x 3 = 12


Product rule for ordered pairs3
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 ways

to choose the first element of

the pair, and n2 ways to choose

the second element, then the

number of possible pairs is

n1n2. Here n1 = 4, n2 = 3.


Counting in either or situations
Counting in “Either-Or” Situations

  • NCAA Basketball Tournament: how many ways can the “bracket” be filled out?

    • How many games?

    • 2 choices for each game

    • 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
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
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
Efficient Methods for Counting Outcomes

  • Factorial Notation:

    n!=12 … n

  • Examples

    1!=1; 2!=12=2; 3!= 123=6; 4!=24;

    5!=120;

  • Special definition: 0!=1


Factorials with calculators and excel
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
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
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
Permutations with calculator and Excel

  • Calculator

    non-graphing: nPr

  • Graphing

    p. 9 of T I Calculator Commands

    (math button)

  • Excel

    Paste: Statistical, Permut


Combinations
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
ST 101 Powerball Lottery

From the numbers 1 through 20,

choose 6 different numbers.

Write them on a piece of paper.



North carolina powerball lottery
North Carolina Powerball Lottery

Prior to Jan. 1, 2009

After Jan. 1, 2009


Visualize your lottery chances
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


Raleigh to ft lauderdale
Raleigh to Ft. Lauderdale…

… still plenty of bills remaining, so continue from …


Ft lauderdale to san diego
… Ft. Lauderdale to San Diego

… still plenty of bills remaining, so continue from…


San diego to seattle
… San Diego to Seattle

… still plenty of bills remaining, so continue from …


Seattle to new york
… Seattle to New York

… still plenty of bills remaining, so continue from …


New york back to raleigh
… New York back to Raleigh

… still plenty of bills remaining, so …


Go around again lay a second path of bills
Go around again! Lay a second path of bills

Still have ~ 5,000 bills left!!


Chances of winning nc powerball lottery
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.




Probability trees
Probability Trees

A Graphical Method for Complicated Probability Problems


Example aids testing
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:

    • 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


Question 1
Question 1

  • What is the probability that a randomly selected person will test positive?


Probability tree approach
Probability Tree Approach

  • A probability tree is a useful way to visualize this problem and to find the desired probability.


Probability tree
Probability Tree

clinical reliability

clinical reliability


Probability tree1
Probability Tree

Multiply

branch probs

clinical reliability

clinical reliability


Question 1 answer
Question 1 Answer

  • What is the probability that a randomly selected person will test positive?

  • P(+) = .00599 + .00994 = .01593


Question 2
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
Question 2 Answer

Answer

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


Summary
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
Recap

  • We have a test with very high clinical reliabilities:

    • 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

  • 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.


Examples1
examples

1. P(A)=.3, P(B)=.4; if A and B are mutually exclusive events, then P(AB)=?

A B = , P(A B) = 0

2. 15 entries in pie baking contest at state fair. Judge must determine 1st, 2nd, 3rd place winners. How many ways can judge make the awards?

15P3 = 2730


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