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    Cheating in Exam

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    Hai------------------------------------------------- 3. a) What is the probability that none of these vehicles requires warranty service? ------------------------------------------------- P(x=0) = 12 C0 (0.10)0 (1-0.10)12-0 ------------------------------------------------- = (1) (1) (0.28243) ------------------------------------------------- =0.28243 ------------------------------------------------- b) What is the probability that exactly nine of these vehicles require warranty service? -------------------------------------------------

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    PROBABILITY P(A U B)= P(A) + P(B) – P(A∩B) If P(A∩B) = 0 then A and B are mutually exclusive and P(AUB) = P(A) + P(B) Joint Probability Marginal Probability PXY(x‚y) = P(X=x ∩ Y=y) PX(x) = ∑P(X=x ∩ Y=y) (For all values of y) Quotient Rule: Multiplication Rule P(A|B) = P(A∩B) / P(B) P(A∩B) = P(A|B) x P(B) = P(B|A) x P(A) Two events are statistically independent if: P(A|B) = P(A) P(B|A) = P(B) P(A∩B) = P(A) P(B) _ _ P(A) = P(A|B)P(B) + P(A|B)P(B) Bayes Rule:

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    Discrete Random Variable

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    |P(X = x) |X(P(X = x) | |0 |0.3 | | |1 |0.2 | | |2 | | | |3 |0.4 | | a. Find the probability that X = 2. b. Find the expected value. Exercise 2 Suppose that you are offered the following “deal.” You roll a die. If you roll a 6‚ you win $10. If you roll a 4 or 5‚ you win $5. If you roll a 1‚ 2‚ or 3‚ you pay $6. a. What

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    Appm 3750 Exam 1 Sp13

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    the honor code pledge printed on your bluebook. No books‚ notes or electronic devices of any kind are allowed. Show all work‚ justify your answers. 1. (25 pts) Suppose events A‚ B and C‚ all defined on the same sample space‚ have the following probabilities: P(A) = 0.22‚ P(B) = 0.25‚ P(C) = 0.28‚ P(A ∩ B) = 0.11‚ P(A ∩ C) = 0.05‚ P(B ∩ C) = 0.07 and P(A ∩ B ∩ C) = 0.01. For each of the following parts‚ your answer should be in the form of a complete mathematical statement. (a) Let D be the event that

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    Freemark Abbey Case

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    As a storm approaches‚ Mr. Jaeger must evaluate the risk of harvesting his Riesling grapes immediately or holding off and taking the chance the grapes become thin or produce no mold and sell at a lower price. Mr. Jaeger must evaluate the risk and the expected revenues related to his different options. The recommendation is that Mr. Jaeger should not harvest the Riesling grapes right now but wait for a better profit given by the possibility of an upcoming rainstorm that may prove to produce a botrytis

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    3232

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    A goes to the first teller‚ B to the second teller‚ and C queues. To standardize the answers‚ let us assume that TA is the length of time in minutes starting from noon until Customer A departs‚ and similarly define TB and TC . (a) What is the probability that Customer A will still be in service at time 12:05? (b) What is the expected length of time that A is in the system? (c) What is the expected length of time that A is in the system if A is still in the system at 12:05? (d) How likely is

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    Triola Statistics Test

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    questions. a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third and fourth guesses are correct. That is‚ find P(WWCC)‚ where C denotes a correct answer and W denotes a wrong answer. b. Make a complete list of the different possible arrangements of 2 wrong answers and 2 correct answers‚ then find the probability for each entry in the list. c. Based on the preceding results‚ what is the probability of getting exactly 2 correct answers when 4 guesses

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    modelling using Random Variables . . . . . . . . . . . . 2 Lecture 2. Random variables. Revision. 2.1 Random Variables and Their Distributions. General. . . . 2.2 Expected value= mean‚ Variance=( Standard Deviation)2 . 2.3 Common models. Specific Probability distributions. . . . . 2.4 Realization of Random Variables. Using Excel . . . . . . . 2.5 Expectation of a function of a random variable . . . . . . . 3 Lecture 3. Multivariate distributions. 3.1 Independence . . . . . . . . . . . . . . 3.2 Covariance

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    Business Modelling

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    Modelling Revision Mutually exclusive event- add the probabilities together to find the probability that one or other of the events will occur. E.g men/woman P(A or B)= P(A)+P(B) Non mutually exclusive- shared characteristic P (A or B)= P(A) + P(B) – P(B+A) Independent events – outcome is known to have no effect on another outcome P (A+B) = P(A) X P(B) Dependant events- outcome of one event affects the probability of the outcome of the other. Probability of the second event said to be dependent on the

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    "Decision Tree" 3 Detailed description of Real Option Technique "First‚ using a decision tree‚ I came up with a simple expected value of $13‚980‚000 based on the costs to complete each phase‚ the probabilities of completing each phase‚ and the costs and probabilities associated with failure at each step in the approval process. The expected value of successful completion with Depression only was $36‚390‚000‚ for weight only $1‚200‚000 and for both $26‚880‚000. The expected value

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