Lect1 dr attia

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basic probability concepts

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Lect1 dr attia

  1. 1. Irrigation and Hydraulics Department Faculty of Engineering – Cairo University Basic Probability Concepts by Dr. Mohamed Attia Assistant Professor Academic Year 2011-2012Probability Experiments A random experiment is an experiment that can result in different outcomes, even though it is repeated in the same manner every time. 1
  2. 2. Probability Experiments An experiment: Flipping a coin once. Rolling a die once. Pulling a card from a deck.Sample Spaces The set of all possible outcomes of a random experiment is called a sample space. You may think of a sample space as the set of all values that a variable may assume. We are going to denote the sample space by S. 2
  3. 3. Examples Experiment: Tossing a coin once. S = {H, T} Experiment: Rolling a die once. S = {1, 2, 3, 4, 5, 6}Examples Experiment: Drawing a card.S = 52 cards in a deck K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 (or Ace) 13 spades (♠), 13 hearts (♥), 13 diamonds (♦) and 13 clubs (♣) 3
  4. 4. Classification of SampleSpaces We distinguish between discrete and continuous sample spaces. The outcomes in discrete sample spaces can be counted. The number of outcomes can be finite or infinite. In the case of continuous sample spaces, the outcomes fill an entire region in the space (interval on the line).EventsAn event, E, is a set of outcomes of a probabilityexperiment, i.e. a subset in the sample space. E ⊆ S.Simple event Outcome from a sample space with one characteristice.g.: A red card from a deck of cardsJoint event Involves two outcomes simultaneously e.g.: An ace that is also red from a deck of cards 4
  5. 5. Visualizing EventsContingency Tables Ace Not Ace Total Black 2 24 26 Red 2 24 26 Total 4 48 52Tree Diagrams Ace Red Full Cards Not an Ace Deck Black Ace of Cards Cards Not an AceVisualizing EventsSometimes it is convenient to represent thesample space by a rectangular region, withevents being circles within the region. Such arepresentation is called Venn diagrams.Venn Diagram A B Sample space S S Events A and B 5
  6. 6. Simple EventsThe Event of a TriangleThere are 5 triangles in this collection of 18 objectsJoint Events The event of a triangle AND red in color Two triangles that are red 6
  7. 7. Special Events Impossible event Null Event e.g.: Club & diamond on one card draw ♣♣ Complement of event For event A, all events not in A Denoted as A’ e.g.: A: queen of diamonds A’: all cards in a deck that are not queen of diamonds Special Events Complement of an event Sample space S A’ A Event A S Complement of A A’ 7
  8. 8. Mutually Exclusive Events Two events, E and F, are called mutually exclusive, if EIF =Φ That is, it is impossible for an outcome to be an occurrence of both events. e.g.: A: queen of diamonds; B: queen of clubs Events A and B are mutually exclusiveCollectively exhaustive events One of the events must occur The set of events covers the whole sample space e.g.: -- A: all the aces; B: all the black cards; C: all the diamonds; D: all the hearts Events A, B, C and D are collectively exhaustive Events B, C and D are also collectively exhaustive 8
  9. 9. Special EventsCollectively Exhaustive Events A, B, C, D are collectively exhaustive A and B are NOT mutually exclusive A and C are NOT mutually exclusive A and D are NOT mutually exclusive B, C, and D are also collectively exhaustive B, C, and D are mutually exclusive A C D B ♦ ♥ Special Events AU B A B S AI B A B S 9
  10. 10. Contingency Table A Deck of 52 CardsRed Ace Ace Not an Total Ace Red 2 24 26 Black 2 24 26 Total 4 48 52 Sample SpaceTree Diagram Event Possibilities Ace Red Cards Not an AceFullDeck of AceCards Black Cards Not an Ace 10
  11. 11. Probability Probability is the numerical measure of the likelihood 1 Certain that an event will occur The probability of an event E is a number, P(E), such that 0 ≤ P( E ) ≤ 1 .5 Sum of the probabilities of all mutually exclusive and collective exhaustive events is 1 0 Impossible P( S ) = 1 Computing Probabilities The probability of an event E: number of event outcomesP( E ) = total number of possible outcomes in the sample space X = T e.g. P( ) = 2/36 (There are 2 ways to get one 6 and the other 4) Each of the outcomes in the sample space is equally likely to occur 11
  12. 12. Computing Joint Probability The probability of a joint event, A and B: P (A and B ) = P (A ∩ B ) number of outcomes from both A and B = total number of possible outcomes in sample space E.g. P (Red Card and Ace) 2 Red Aces 1 = = 52 Total Number of Cards 26Joint Probability UsingContingency Table Event Event B1 B2 Total A1 P(A1 and B1) P(A1 and B2) P(A1) A2 P(A2 and B1) P(A2 and B2) P(A2) Total P(B1) P(B2) 1 Joint Probability Marginal (Simple) Probability 12
  13. 13. Computing Compound (orMultiple) Probability Probability of a compound event, A or B: P( A or B ) = P( A ∪ B ) number of outcomes from either A or B or both = total number of outcomes in sample space E.g. P (Red Card or Ace) 4 Aces + 26 Red Cards - 2 Red Aces = 52 total number of cards 28 7 = = 52 13Compound Probability(Addition Rule) P(A1 or B1 ) = P(A1) + P(B1) - P(A1 and B1) Event Event B1 B2 Total A1 P(A1 and B1) P(A1 and B2) P(A1) A2 P(A2 and B1) P(A2 and B2) P(A2) Total P(B1) P(B2) 1For Mutually Exclusive Events: P(A or B) = P(A) + P(B) 13
  14. 14. Computing Conditional Probability The probability of event A given that event B has occurred: P ( A and B ) A B P( A | B) = P( B) A&B It means the likelihood of realizing a sample point in A assuming that it belongs to B E.g. P(Red Card given that it is an Ace) 2 Red Aces 1/52 = = = 0.5 4 Aces 2 /52 Conditional Probability Using Contingency Table Color Type Red Black Total Ace 2 2 4 Non-Ace 24 24 48 Total 26 26 52 Revised Sample Space P (Ace and Red) 2 / 52 2P (Ace | Red) = = = P(Red) 26 / 52 26 14
  15. 15. Conditional Probability andStatistical Independence Conditional probability: P ( A and B) P( A | B) = P( B) Multiplication rule: P ( A and B) = P ( A | B ) P ( B ) = P( B | A) P ( A)Conditional Probability andStatistical Independence (continued) Events A and B are independent if P( A | B) = P ( A) or P( B | A) = P ( B ) or P( A and B) = P( A) P( B) Events A and B are independent when the probability of one event, A, is not affected by another event, B 15
  16. 16. Bayes’s Theorem B1, B2, …, Bk are mutually exclusive and collectively exhaustive Suppose it is known that event A has occurred What is the (conditional) probability that event Bi has occurred? B1 B2 Bk A Bayes’s Theorem P ( A | Bi ) P ( Bi )P ( Bi | A ) = P ( A | B1 ) P ( B1 ) + • • • + P ( A | Bk ) P ( Bk ) P ( Bi and A ) = P ( A) Adding up the parts Same of A in all Event the B’s 16
  17. 17. Bayes’s Theorem Using ContingencyTable (Example 1)Fifty percent of graduating engineers in a certain yearworked as water resources (WR) managers. Out of thosewho worked as WR managers, 40% had a Masters degree.Ten percent of those graduating engineers who did not workas WR managers had a Masters degree. What is theprobability that a randomly selected engineer who has aMasters degree is a WR manager?P (WR) = 0.5 P( M / WR ) = 0.4 P( M / W R ) = 0.1P (WR / M ) = ?Bayes’s TheoremUsing Contingency Table (continued) WR WR Total Masters .2 .05 .25 Masters .3 .45 .75 Total .5 .5 1.0 P( M | WR ) P (WR ) P (WR / M ) = P( M | WR ) P (WR ) + P( M | W R ) P(W R ) (0.4)(0.5) 0.2 = = = 0.8 (0.4)(0.5) + (0.1)(0.5) 0.25 17
  18. 18. Example 2A water treatment plant may fail for two reasons: inadequacyof materials (event A) or mechanical failure (event B).If P(A) = 2 P(B), P(A | B) = 0.8 and the probability of failure ofthe treatment plant equals 0.001 , what is the probability thata mechanical failure occurs? What is the probability of afailure due to inadequate materials?The probability of failure P (A U B) may be written as: P (A U B) = P (A) + P (B) – P (A ∩ B) = 2 P (B) + P (B) – P (B) P (A | B) = 3 P (B) – 0.8P (B) 0.001 = 3 P (B) – 0.8P (B) P (B) = 0.00045 P (A) = 0.0009Example 3 Each of two pumps Q1 and Q2 may not operate. On inspection one may observe one of the outcomes of the following set: {(ƒ, ƒ), (ƒ, o), (o, ƒ), (o, o)}. The notation (ƒ, o) means that pump Q1 fails and pump Q2 operates. The sample space has four elements. From experience one knows that: P ((ƒ, ƒ)) = 0.1, P ((ƒ, o )) = 0.2 P ((o, ƒ)) = 0.3, P ((o, o )) = 0.4 Let A: Q1 operates B: Q2 operates C: at least one of the pumps operates Compute P(A), P(B), P(A ∩ B), P(C) 18
  19. 19. Example 3 (Cont.) P ( A ) = 0.7 P ( B ) = 0.6 P ( A ∩ B ) = P (( o , o )) = 0.4 P(C)=(AUB) = P ( A ) + P ( B ) – P ( A ∩ B ) = 0.9Example 4 The demand of a water supply system can be low (event L) , moderate (event M) or high (event H) with known probabilities P(L) = 0.1, P(M) = 0.7 and P(H) = 0.2. Failure of the system (event F) can only occur if a certain pump fails to function. From past experience the following conditional probabilities are known: P(F | L) = 0.05, P(F | M) = 0.10 and P(F | H) = 0.25. What is the probability of a pump failure P(F). First we observe that the events L, M and H are clearly disjoint and that their union is just the sure event S. Hence P ( F ) = P (( L ∩ F ) U ( M ∩ F ) U ( H ∩ F )) = P (L ∩ F) + P (M ∩ F) + P (H ∩ F) Using the definition of conditional probability, we get P ( F ) = P (L) P (F | L) + P (M) P (F | M) + P (H) P (F | H) = 0.1 × 0.05 + 0.7 × 0.10 + 0.2 × 0.25 = 0.125 19
  20. 20. Example 5 Weather forecast information is sent using one of four routes. Let Ri denote the event that route i is used for sending a message (i = 1, 2, 3, 4). The probabilities of using the four routes are 0.1, 0.2, 0.3 and 0.4, respectively. Further, it is known that the probabilities of an error being introduced while transmitting messages are 0.10, 0.15, 0.20 and 0.25 for routes 1, 2, 3 and 4, respectively. In sending a message an error occurred. What is the probability that route 2 was used for transmission? Let E denote the event of having an erroneous message P(E | R1) = 0.10, P(E | R2) = 0.15, P(E | R3) = 0.20, P(E | R4) = 0.25.Example 5 (Cont.) Using Bayes rule one can compute the probability that the message was sent via route 2 given the event that an error has occurred. This probability is given by P( R2 I E ) P( R2 ) P( E | R2 ) P(R2 | E) = = 4 P( E ) ∑ i =1 P( Ri ) P( E | Ri ) 0.2 × 0.15 = 0.1× 0.10 + 0.2 × 0.15 + 0.3 × 0.20 + 0.4 × 0.25 0.03 = = 0.15. 0.20 20
  21. 21. Example 6 The probability of receiving more than 50 mm of rain in the months of the year is 0.25, 0.30, 0.35, 0.40, 0.20, 0.10, 0.05, 0.05, 0.05, 0.05, 0.10 and 0.20 for January, February, ..., December, respectively. A monthly rainfall record selected at random is found to be more than 50 mm. What is the probability that this record belongs to month m (m = 1 for January, ..., m = 12 for December). Denote the probability of selecting month m as P(Mm). Denote the conditional probability of receiving more than 50 mm of rain, given that month m is selected as P(E | Mm). Then P(Mm | E) can be computed using Bayes rule as indicated in the following equation and table P( M j ) P( E | M j ) P(M j ) P( E | M j ) P( M j | E ) = = 12 P( E ) ∑ P(M m =1 m ) P( E | M m )Example 6 (Cont.) Application of Bayes’ rule m P(Mm) P(E | Mm) P(Mm) P(E | Mm) P(Mm | E) 1 1/12 0.25 0.02083 0.1191 2 1/12 0.30 0.02500 0.1429 3 1/12 0.35 0.02917 0.1667 4 1/12 0.40 0.03333 0.1905 5 1/12 0.20 0.01667 0.0952 6 1/12 0.10 0.00833 0.0476 7 1/12 0.05 0.00417 0.0238 8 1/12 0.05 0.00417 0.0238 9 1/12 0.05 0.00417 0.0238 10 1/12 0.05 0.00417 0.0238 11 1/12 0.10 0.00833 0.0476 12 1/12 0.20 0.01667 0.0952 12/12 0.17500 = P(E) 1.0000 21
  22. 22. Thank you 22

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