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Probability

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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
• 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 IV
                           out of V times
     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.)
• Subjective probability
  individual assigns prob. based on personal
  experience, anecdotal evidence, etc.
• Classical approach
  every 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 coin
2. Draw 1 card from a standard deck of cards
3. 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 outcome
  Null event: the empty set F
  Certain event: S
                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


1. 0  P ( A)  1, for any event A

2. P (F )  0, P ( S )  1
        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
• 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: 23
No. of people    23 30 40 60
Probability      .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}
                            364   363
  3 people     :P ( A')        
                            365   365
   23 people    :
              364    363          343
   P ( A')                            . 498
              365    365          365
   so P ( A )  1  P ( A ' )  1  . 498  . 502
    Unions and Intersections


S               A

        A             B




              A
     Mutually Exclusive Events
• Mutually exclusive events-no outcomes
  from S in common
                  A = 
 S
          A
                       B
     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)
• 5. For two independent events A and B
            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)
    P(AB)=P(A) + P(B) - P(A B)


S                 A

          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, 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



    The Equally Likely Approach
(also called the Classical Approach)
      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
We Need Efficient Methods for
Counting Outcomes
  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,4cjunior colleges
• possible pairs: 3 state colleges
                      total number of possible
(1, a) (1, b) (1, c)  pairs = 4 x 3 = 12
(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
                         In general, if there are n1 ways
• state colleges a, b, cthe pair, and n ways to choose
                         to choose the first element of
                                         2
• possible pairs:        the second element, then the
                         number of possible pairs is
(1, a) (1, b) (1, c)     n1n2. Here n1 = 4, n2 = 3.
(2, a) (2, b) (2, c)
(3, a) (3, b) (3, c)
(4, a) (4, b) (4, c)
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 game
    3. 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!=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
• 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
                   ABCDE
• 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 (cont.)

                5!       5!
5  4  20             5 4
             (5  2)! 3!
                      5!
Notation : 5 P2             20
                   (5  2)!
  Permutations with calculator
           and Excel
• Calculator
  non-graphing: nPr
• Graphing
  p. 9 of T I Calculator Commands
  (math button)
• Excel
  Paste: Statistical, Permut
            Combinations
                    ABCDE
• 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
                Combinations (cont.)

                           5! 5  4 20
 
 5
 2     5 C2 
                   5!
                                 
               (5  2)!2! 3!2! 1 2 2
                                       10



 
 n
 r     n Cr 
                   n!
               (n  r )!r!
ST 101 Powerball Lottery



       From the numbers 1 through 20,
         choose 6 different numbers.
       Write them on a piece of paper.
         Chances of Winning?
Choose 6 numbers from 20, without
replacement, order not important.
Number of possibilit ies?

 
 20
 6     20 C6 
                   20!
                (20  6)!6!
                             38,760
  North Carolina Powerball Lottery
Prior to Jan. 1, 2009         After Jan. 1, 2009
  5 from 1 - 55:              5 from 1 - 59:
   55!                         59!
          3, 478, 761                5, 006, 386
  5!50!                       5!54!
  1 from 1 - 42 (p'ball #):   1 from 1 - 39 (p'ball #):
   42!                         39!
         42                         39
  1!41!                       1!38!
  3, 478, 761*42             5, 006, 386*39 
  146,107, 962
                              195, 249, 054
 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…




                   … still plenty of bills
                   remaining, so continue
                   from …
         … Ft. Lauderdale to San Diego




… still plenty of bills remaining, so continue from…
… San Diego to Seattle




  … still plenty of bills remaining, so continue from …
… Seattle to New York




 … still plenty of bills remaining, so continue from …
    … New York back to Raleigh




… still plenty of bills remaining, so …
Go around again! Lay a second
        path of bills




Still have ~ 5,000 bills left!!
      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
Choose 6 numbers from 54 numbers without
replacement; order not important
         54!
54 C6         25,827,165
        48!6!
(about 1 secondin 10 months)
(1200 ft 2 house,16.5 million ping pong balls)
        Virginia State Lottery

                  50!
Pick 5 : 50 C5         2,118,760
                 45!5!

2,118,760  25 C1 
            25!
2,118,760         52,969000
            24!1!
       Probability Trees



A Graphical Method for Complicated
       Probability Problems
                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
               Question 1
• What is the probability that a randomly
  selected person will test positive?
     Probability Tree Approach


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




clinical
reliability
                Probability Tree
  clinical                           Multiply
  reliability                      branch probs




clinical
reliability
          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
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
• 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 .999
    2.   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.
                        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