Probability

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Name                                       Shakeel Nouman
Religion                                  Christian
Domicile                            Punjab (Lahore)
Contact #             0332-4462527. 0321-9898767
E.Mail                                sn_gcu@yahoo.com
sn_gcu@hotmail.com

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Probability

  1. 1. Probability Shakeel Nouman M.Phil Statistics Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 1
  2. 2. 2 Slide 2 Probability Using Statistics  Basic Definitions: Events, Sample Space, and Probabilities  Basic Rules for Probability  Conditional Probability  Independence of Events  Combinatorial Concepts  The Law of Total Probability and Bayes’ Theorem  Summary and Review of Terms  Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  3. 3. 2-1 Probability is:     Slide 3 A quantitative measure of uncertainty A measure of the strength of belief in the occurrence of an uncertain event A measure of the degree of chance or likelihood of occurrence of an uncertain event Measured by a number between 0 and 1 (or between 0% and 100%) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  4. 4. Types of Probability  Objective or Classical Probability  based on equally-likely events  based on long-run relative frequency of events  not based on personal beliefs  is the same for all observers (objective)  examples: toss a coin, throw a die, pick a card Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 4
  5. 5. Types of Probability (Continued)  Slide 5 Subjective Probability  based on personal beliefs, experiences, prejudices, intuition personal judgment  different for all observers (subjective)  examples: Super Bowl, elections, new product introduction, snowfall Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  6. 6. 2-2 Basic Definitions  Slide 6 Set - a collection of elements or objects of interest  Empty set (denoted by )  a set containing no elements  Universal set (denoted by S)  a set containing all possible elements  Complement (Not). The complement of A is  a set containing all elements of S not in A  A Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  7. 7. Complement of a Set Slide 7 S A A Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  8. 8. Basic Definitions (Continued) Slide 8 Intersection (And)  B A – a set containing all elements in both A and B Union (Or) –  A  B a set containing all elements in A or B or both Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  9. 9. Sets: A Intersecting with B Slide 9 S A B A B Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  10. 10. Sets: A Union B Slide 10 S A B A B Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  11. 11. Basic Definitions (Continued) Slide 11 • Mutually exclusive or disjoint sets –sets having no elements in common, having no intersection, whose intersection is the empty set • Partition –a collection of mutually exclusive sets which together include all possible elements, whose union is the universal set Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  12. 12. Mutually Exclusive or Disjoint Sets Slide 12 Sets have nothing in common S A B Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  13. 13. Sets: Partition Slide 13 A3 A1 A2 A4 A5 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer S
  14. 14. Experiment Slide 14 • Process that leads to one of several possible outcomes *, e.g.: Coin toss » Heads,Tails Throw die » 1, 2, 3, 4, 5, 6 Pick a card » AH, KH, QH, ... • • Introduce a new product Each trial of an experiment has a single observed outcome. The precise outcome of a random experiment is unknown before a trial.  Also called a basic outcome elementary event or simple event Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  15. 15. Events : Definition  Slide 15 Sample Space or Event Set Set of all possible outcomes (universal set) for a given experiment  E.g.: Throw die • S = {1,2,3,4,5,6}  Event Collection of outcomes having a common characteristic  E.g.: Even number • A = {2,4,6} – Event A occurs if an outcome in the set A occurs  Probability of an event Sum of the probabilities of the outcomes of which it consists  P(A) = P(2) + P(4) + P(6) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  16. 16. Equally-likely Probabilities (Hypothetical or Ideal Experiments) Slide 16 • For example: Throw a die » Six possible outcomes {1,2,3,4,5,6} » If each is equally-likely, the probability of each is 1/6 = .1667 = 16.67% 1 P ( e)  » n( S ) » Probability of each equally-likely outcome is 1 over the number of possible outcomes  Event A (even number) » P(A) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 1/2 » P( A)   P( e) for e in A n( A ) 3 1    n( S ) 6 2 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  17. 17. Pick a Card: Sample Space Hearts Diamonds Clubs Spades A K Q J 10 9 8 7 6 5 4 3 2 A K Q J 10 9 8 7 6 5 4 3 2 A K Q J 10 9 8 7 6 5 4 3 2 A Union of K vents eart Q and Ace J 10 9 8 7 6 5 4 3 2 P ( Heart  Ace )  n ( Heart  Ace )  n(S ) 16 4  52 13 vent eart n ( Heart ) P ( Heart )  13  n(S ) 1  52 Slide 17 vent Ace n ( Ace ) P ( Ace )  4 1  n(S )  52 13 The intersection of the events eart and Ace comprises the single point circled twice: the ace of hearts 4 n ( Heart  Ace ) P ( Heart  Ace )  1  n(S ) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer 52
  18. 18. 2-3 Basic Rules for Probability  Range of Values0  P( A)  1  Slide 18 Complements - Probability of not A P( A)  1 P( A)  Intersection - Probability of both A and B n( A  B) P( A  B)  n( S ) Mutually exclusive events (A and C) : P( A  C )  0 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  19. 19. Basic Rules for Probability (Continued) • Slide 19 Union - Probability of A or B or both (rule of unions) P( A  B)  n( A  B)  P( A)  P( B)  P( A  B) n( S ) Mutually exclusive events: If A and B are mutually exclusive, then P( A  B)  0 so P( A  B)  P( A)  P( B) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  20. 20. Sets: P(A Union B) Slide 20 S A B P( A  B) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  21. 21. Basic Rules for Probability (Continued) Slide 21 • Conditional Probability - Probability of A given B P( A B)  P( A  B) , where P( B)  0 P( B) Independent events: P( A B)  P( A) P( B A)  P( B) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  22. 22. 2-4 Conditional Probability Rules of conditional probability: P( A B)  P( A  B) so P( A  B)  P( A B) P( B) P( B)  P( B A) P( A) If events A and D are statistically independent: P ( A D )  P ( A) so P( A  D)  P( A) P( D) P ( D A)  P ( D ) Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 22
  23. 23. Contingency Table - Example 22 Slide 23 Counts AT& T IBM Total Telecommunication 40 10 50 Computers 20 30 50 Total 60 40 100 Probabilities AT& T IBM Total Telecommunication .40 .10 .50 Computers .20 .30 .50 Total .60 .40 Probability that a project is undertaken by IBM given it is a telecommunications project: P ( IBM  T ) P(T ) .10  .2 .50 P ( IBM T )  1.00 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  24. 24. 2-5 Independence of Events Slide 24 Conditions for the statistical independence of events A and B: P ( A B )  P ( A) P ( B A)  P ( B ) and P ( A  B )  P ( A) P ( B ) P ( Ace  Heart ) P ( Heart ) 1 1  52   P ( Ace ) 13 13 52 P ( Ace Heart )  P ( Ace  Heart )  P ( Heart  Ace ) P ( Ace ) 1 1  52   P ( Heart ) 4 4 52 P ( Heart Ace )  4 13 1   P ( Ace ) P ( Heart ) 52 52 52 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  25. 25. Independence of Events Example 2-5 Events Television (T) and Billboard (B) are assumed to be independent. a) P ( T  B )  P ( T ) P ( B )  0.04 * 0.06  0.0024 b) P ( T  B )  P ( T )  P ( B )  P ( T  B )  0.04  0.06  0.0024  0.0976 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 25
  26. 26. Product Rules for Independent Events Slide 26 The probability of the intersection of several independent events is the product of their separate individual probabilities: P( A  A  A  An )  P( A ) P( A ) P( A ) P( An ) 1 2 3 1 2 3 The probability of the union of several independent events is 1 minus the product of probabilities of their complements: P( A  A  A  An )  1 P( A ) P( A ) P( A ) P( An ) 1 2 3 1 2 3 Example 2-7: (Q  Q  Q Q )  1  P(Q ) P(Q ) P(Q ) P(Q ) 1 2 3 10 1 2 3 10  1.9010  1.3487 .6513 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  27. 27. 2-6 Combinatorial Concepts Slide 27 Consider a pair of six-sided dice. There are six possible outcomes from throwing the first die {1,2,3,4,5,6} and six possible outcomes from throwing the second die {1,2,3,4,5,6}. Altogether, there are 6*6=36 possible outcomes from throwing the two dice. In general, if there are n events and the event i can happen in Ni possible ways, then the number of ways in which the sequence of n events may occur is N1N2...Nn.  Pick 5 cards from a deck of 52 with replacement  52*52*52*52*52=525 380,204,032 different possible outcomes  Pick 5 cards from a deck of 52 without replacement  52*51*50*49*48 = 311,875,200 different possible outcomes Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  28. 28. More on Combinatorial Concepts (Tree Diagram)  . . . . . . . .  Order the letters: A, B, and C C B C B A C C A B C A B A B A  . . . .  ABC ACB BAC BCA CAB CBA Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 28
  29. 29. Factorial Slide 29 How many ways can you order the 3 letters A, B, and C? There are 3 choices for the first letter, 2 for the second, and 1 for the last, so there are 3*2*1 = 6 possible ways to order the three letters A, B, and C. How many ways are there to order the 6 letters A, B, C, D, E, and F? (6*5*4*3*2*1 = 720) Factorial: For any positive integer n, we define n factorial as: n(n-1)(n-2)...(1). We denote n factorial as n!. The number n! is the number of ways in which n objects can be ordered. By definition 1! = 1 and 0! = 1. Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  30. 30. Permutations (Order is important) Slide 30 What if we chose only 3 out of the 6 letters A, B, C, D, E, and F? There are 6 ways to choose the first letter, 5 ways to choose the second letter, and 4 ways to choose the third letter (leaving 3 letters unchosen). That makes 6*5*4=120 possible orderings or permutations. Permutations are the possible ordered selections of r objects out of a total of n objects. The number of permutations of n objects taken r at a time is denoted by nPr, where Pr  n! n ( n  r )! 6 For example: 6! 6! 6 * 5 * 4 * 3 * 2 * 1 P    6 * 5 * 4  120 (6  3)! 3! 3 * 2 *1 3 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  31. 31. Combinations (Order is not Important) Slide 31 Suppose that when we pick 3 letters out of the 6 letters A, B, C, D, E, and F we chose BCD, or BDC, or CBD, or CDB, or DBC, or DCB. (These are the 6 (3!) permutations or orderings of the 3 letters B, C, and D.) But these are orderings of the same combination of 3 letters. How many combinations of 6 different letters, taking 3 at a time, are there? n Combinations are the possible selections of r items from a group of n items     regardless of the order of selection. The number of combinations is denotedr and is read as n choose r. An alternative notation is nCr. We define the number of combinations of r out of n elements as: n!  n  n Cr     r r!(n  r)! For example: 6! 6! 6 * 5 * 4 * 3 * 2 * 1 6 * 5 * 4 120  n      20    6 C3   r 3!(6  3)! 3!3! (3 * 2 * 1)(3 * 2 * 1) 3 * 2 * 1 6 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  32. 32. Example: Template for Calculating Permutations & Combinations Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 32
  33. 33. 2-7 The Law of Total Probability and Bayes’ Theorem The law of total probability: P( A)  P( A  B)  P( A  B ) In terms of conditional probabilities: P( A)  P( A  B)  P( A  B )  P( A B) P( B)  P( A B ) P( B ) More generally (where Bi make up a partition): P( A)   P( A  B ) i   P( AB ) P( B ) i i Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 33
  34. 34. The Law of Total ProbabilityExample 2-9 Slide 34 Event U: Stock market will go up in the next year Event W: Economy will do well in the next year P(U W ) .75 P(U W )  30 P(W ) .80  P(W )  1.8 .2 P(U )  P(U W )  P(U W )  P(U W ) P(W )  P(U W ) P(W )  (.75)(.80)  (.30)(.20) .60.06 .66 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  35. 35. Bayes’ Theorem Slide 35 • Bayes’ theorem enables you, knowing just a • little more than the probability of A given B, to find the probability of B given A. Based on the definition of conditional probability and the law of total probability. P ( A  B) P ( A) P ( A  B)  P ( A  B)  P ( A  B ) P ( A B) P ( B)  P ( A B ) P ( B)  P ( A B ) P ( B ) P ( B A)  Applying the law of total probability to the denominator Applying the definition of conditional probability throughout Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  36. 36. Bayes’ Theorem - Example 2-10 Slide 36 • A medical test for a rare disease (affecting 0.1% of the population [ P( I )  0.001 ]) is imperfect: When administered to an ill person, the test will indicate so with probability 0.92 [ P(Z I )  .92  P(Z I )  .08 ] » The event (Z I ) is a false negative When administered to a person who is not ill, the test will erroneously give a positive result (false positive) with probability 0.04 [ P(Z I )  0.04  P(Z I )  0.96 ] » The event (Z I ) is a false positive. . Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  37. 37. Example 2-10 (continued) P ( I )  0.001 P ( I )  0.999 P ( Z I )  0.92 P ( Z I )  0.04 Slide 37 P( I  Z ) P( Z ) P( I  Z )  P( I  Z )  P( I  Z ) P( Z I ) P( I )  P( Z I ) P( I )  P( Z I ) P( I ) P( I Z )  (.92)( 0.001) (.92)( 0.001)  ( 0.04)(.999) 0.00092 0.00092   0.00092  0.03996 .04088 .0225  Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  38. 38. Example 2-10 (Tree Diagram) Prior Probabilities Conditional Probabilities P( Z I )  092 . P( I )  0.001 P( I )  0999 . P( Z I )  008 . P( Z I )  004 . Joint Probabilities P( Z  I )  (0.001)(0.92) .00092 P( Z  I )  (0.001)(0.08) .00008 P( Z  I )  (0.999)(0.04) .03996 P( Z I )  096 . P( Z  I )  (0.999)(0.96) .95904 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 38
  39. 39. Bayes’ Theorem Extended • Slide 39 Given a partition of events B1,B2 ,...,Bn: P( A  B ) P( B A)  P( A) P( A  B )   P( A  B ) P( A B ) P( B )   P( A B ) P( B ) 1 1 Applying the law of total probability to the denominator 1 i 1 Applying the definition of conditional probability throughout 1 i i Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  40. 40. Bayes’ Theorem Extended Example 2-11    Slide 40 An economist believes that during periods of high economic growth, the U.S. dollar appreciates with probability 0.70; in periods of moderate economic growth, the dollar appreciates with probability 0.40; and during periods of low economic growth, the dollar appreciates with probability 0.20. During any period of time, the probability of high economic growth is 0.30, the probability of moderate economic growth is 0.50, and the probability of low economic growth is 0.50. Suppose the dollar has been appreciating during the present period. What is the probability we are experiencing a period of high economic growth? Partition: H - High growth P(H) = 0.30 M - Moderate growth P(M) = 0.50 L - Low growth P(L) = 0.20 Event A  Appreciation P ( A H )  0.70 P ( A M )  0.40 P ( A L)  0.20 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  41. 41. Example 2-11 (continued) P ( H  A) P ( H A)  P ( A) P ( H  A)  P ( H  A)  P ( M  A)  P ( L  A) P( A H ) P( H )  P ( A H ) P ( H )  P ( A M ) P ( M )  P ( A L) P ( L) ( 0.70)( 0.30)  ( 0.70)( 0.30)  ( 0.40)( 0.50)  ( 0.20)( 0.20) 0.21 0.21   0.21 0.20  0.04 0.45  0.467 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 41
  42. 42. Example 2-11 (Tree Diagram) Prior Probabilities Conditional Probabilities P ( A H )  0.70 P ( H )  0.30 P ( A H )  0.30 P ( A M )  0.40 Joint Probabilities P ( A  H )  ( 0.30)( 0.70)  0.21 P ( A  H )  ( 0.30)( 0.30)  0.09 P ( A  M )  ( 0.50)( 0.40)  0.20 P ( M )  0.50 P ( A M )  0.60 P ( A  M )  ( 0.50)( 0.60)  0.30 P ( L )  0.20 P ( A L )  0.20 P ( A L )  0.80 Slide 42 P ( A  L )  ( 0.20)( 0.20)  0.04 P ( A  L )  ( 0.20)( 0.80)  0.16 Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
  43. 43. 2-8 Using Computer: Template for Calculating the Probability of at least one success Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer Slide 43
  44. 44. Slide 44 Name Religion Domicile Contact # E.Mail M.Phil (Statistics) Shakeel Nouman Christian Punjab (Lahore) 0332-4462527. 0321-9898767 sn_gcu@yahoo.com sn_gcu@hotmail.com GC University, . (Degree awarded by GC University) M.Sc (Statistics) Statitical Officer (BS-17) (Economics & Marketing Division) GC University, . (Degree awarded by GC University) Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development Department, Govt. of Punjab Probability By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer

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