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PROBABILITY MODELS

PROBABILITY MODELS. 1.1 Probability Models and Engineering. Probability models are applied in all aspects of Engineering. Traffic engineering, reliability, manufacturing process control, design of industrial experiments, signal processing, decision analysis and risk analysis.

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PROBABILITY MODELS

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  1. PROBABILITY MODELS

  2. 1.1 Probability Models and Engineering Probability models are applied in all aspects of Engineering • Traffic engineering, reliability, manufacturing process control, design of industrial experiments, signal processing, decision analysis and risk analysis • Measurements in every aspect of engineering are subject to variability. Engineers deal with measurement data every day and it is vital for them to understand the relevance of measurement error. • Probability methods are used to model variability and uncertainty • Quality in manufacturing is inversely proportional to the variability of the manufacturing process

  3. At the end of a TV game show (Let’s Make a Deal) the winning contestant selects one of 3 curtains. Behind one curtain are the keys to a new car. The spaces behind the other curtains are empty. When the choice is made it has always been the practice of the host to open one of the other curtains to reveal an empty space. The contestant is then offered the chance to change their mind. Does opening the other curtain make a difference? Should the contestant change their selection? Example 1: Let’s Make a Deal Selected Opened

  4. A B C A B C A B C One of these closed curtains hides the car keys You choose A Host opens C YES Should you switch your choice to B?

  5. A C E 2 1 B D Example 2: Reliability of a Network A 2 terminal network has 6 components A, B, C, D and E connected as follows: Each component has reliability 0.9 over the time period. What is the reliability of the connection between 1 & 2? To solve this example we will need to consider conditional probabilities

  6. 1.3 Building a Probability Model For a simple situation involving a Random Experiment A probability model is built up by following the steps: 1. List all the possible elementary outcomes This list is called the sample space of the experiment. 2. Assign probability weights to each elementary outcome. 3. Identify events, as subsets of the sample space, for which probabilities are to be calculated. The probability that an event A occurs is the sum of the probability weights of the elementary outcomes comprising A. This is denoted by P(A).

  7. P(A) is a measure of the likelihood that an event A will occur when the experiment is performed. Building a probability model this way guarantees that the probabilities of all events conform to the 3 basic laws of probability: • P(A)  0, for every event A. • P(S) = 1, where S is the sample space. Equivalent to saying the list is complete • P(either A or B occurs) = P(A)  P(B), for any pair of disjoint events A and B. Events A and B disjoint means that they cannot occur together for the same experiment.

  8. Two dice are rolled. (As in Backgammon, Monopoly or Craps) Example 3: Two Dice Sample space is: S ={(1,1), (1,2), (1,3), (1,4), (1,5) (1,6), (2,1), (2,2), (2,3), (2,4), (2,5) (2,6), (3,1), (3,2), (3,3), (3,4), (3,5) (3,6), (4,1), (4,2), (4,3), (4,4), (4,5) (4,6), (5,1), (5,2), (5,3), (5,4), (5,5) (5,6), (6,1), (6,2), (6,3), (6,4), (6,5) (6,6)} ={(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. A = total score is 7 B = both dice even = {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2),(6,4), (6,6)} C = at least one 6 = {(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2), (6,3), (6,4), (6,5)} D = doubles = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

  9. A  B is the event that both A and B occur. S S B A B A then A and B cannot occur together for the same experiment. NOTE: If A  B =  1.4 Combining Events IF A and B are events then: A and B are mutually exclusive or disjoint events.

  10. VENN DIAGRAMS

  11. Do the ( ) first

  12. PROBABILITY

  13. A probability is a number assigned to an event representing the chance or likelihood that the event occurs when the random experiment is performed. The probability of an event A is denoted P(A) Probabilities have to be assigned in a consistent way. The probabilities of all events of a random experiment must satisfy the three rules 1 P(A)0 for any event A2 P(S) = 13 P(AB) = P(A) + P(B) for any pair of disjoint events

  14. Result: Complement: P(A ) = 1 - P(A) A S P(A)=p P(S)=1 A P(A)=1-p

  15. A B P(A) P(B) Notice that when we add the probabilities, this region, is added in twice - once from A and once from B Result: Union: P(AB ) = P(A) + P(B) - P(A B) S So we subtract P (A B) to correct the double overlap

  16. THREE IMPORTANT RULES Complement: P(A ) = 1 - P(A) Intersection P(AB) = P(A).P(B) If A and B are independent events. (More on independence later) Union: P(AB ) = P(A) + P(B) - P(A B) Note:  and  multiply  or  add

  17. Example A and B are independent events with P(A) = 0.7 P(B) = 0.6 Find P(A B) P(A B) = P(A) + P(B) - P(AB) so we need P(AB) As A,B are independent P(AB) = P(A).P(B) = 0.42 P(A B) = 0.7 + 0.6 - 0.42 = 0.88

  18. MORE COMPLICTED EVENTS

  19. 0.3 0.2 • If P(A) = 0.7, P(B) = 0.6 and P(A  B) = 0.9, • determine P(A B) Example Strategy: Find the probability of each of the disjoint areas in the Venn Diagram P(AB ) = P(A) + P(B) - P(A B) 0.9 = 0.7 + 0.6 - P(A B) P(A B) = 0.4 0.4 0.1 Region outside A,B has probability 1 - (0.3 +0.4 +0.2) = 0.1

  20. B A S 0.3 0.4 0.2 0.1 P(A B) ie outside A or in B

  21. B A S 0.3 0.4 0.2 0.1 P(A B) ie outside A or in B P(A B) = 0.1 + 0.4 + 0.2

  22. Example A cup coffee which is supposed to contain milk and sugar obtained from the coffee dispensing machine in an engineering school cafeteria is likely to have a number of different short-comings. They are represented as the events: A - coffee burnt, B - no sugar, C - no milk. It is known that: P(A) = 0.7, P(B) = 0.4 and P(A  B) = 0.2, P(C) = 0.3 P(A  C) = 0.2, P(B  C) = 0.2 , P(AB C) = 0.1 Calculate the probability that the coffee: (1) is burnt but has sugar and milk, (2) is not burnt and either has no sugar or has milk.

  23. P(A) = 0.7, P(B) = 0.4 and P(A  B) = 0.2, P(C) = 0.3 P(A  C) = 0.2, P(B  C) = 0.2 , P(AB C) = 0.1 A Good Strategy: Work out from the centre 0.1 0.4 0.1 0.1 0.1 0.1 0.0 0.1

  24. 0.1 0.4 0.4 0.1 0.1 0.1 0.1 0.0 0.1 A - coffee burnt, B - no sugar, C - no milk. PROBLEM (1) What corresponds to: is burnt but has sugar and milk, Translate: Burnt and Sugar and Milk AB  C In A and not in B and not in C Answer = 0.4

  25. 0.1 0.4 0.4 0.1 0.1 0.1 0.1 0.0 0.1 A - coffee burnt, B - no sugar, C - no milk. PROBLEM 2: What corresponds to: is not burnt and either has no sugar or has milk Translate: Not Burnt and (No Sugar or Milk) A (B  C) Not in A and (in B or outside C)

  26. 0.1 0.4 0.1 0.1 0.1 0.1 0.0 0.1 A - coffee burnt, B - no sugar, C - no milk. What corresponds to: (2) is not burnt and either has no sugar or has milk Translate: Not Burnt and (No Sugar or Milk) A (B  C) Not in A and (in B or outside C) Answer = 0.3

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