Introduction to Probability - Chapter 10 pptx

Introduction to Probability - Chapter 10 pptx

Introduction to Probability - Chapter 10 pptx

... -5 0 00000000000 0-1 00 111111249797300 233420000000 0 100 000000000 -5 0 2100 00000000 0 3347111714111 1101 6251300 00000000000 0-1 00 12211 3100 000 -5 0 00000000000 0-1 00 2 3100 0000000 0 3100 00000000 50 100 000000000 -5 0 344 7101 191112141 310 550 13349579886 3100 104 66 9101 30000 ... follows: 365 392 CHAPTER 10. GENERATING FUNCTIONS Z 1 Z 2 Z 3 Z 4 Z 5 Z...

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Introduction to Probability - Chapter 7 pptx

Introduction to Probability - Chapter 7 pptx

... X 3 .Thus P (S 3 =3) = P(S 2 =2)P (X 3 =1) 300 CHAPTER 7. SUMS OF RANDOM VARIABLES -1 5 -1 0 -5 5 10 15 0.025 0.05 0.075 0.1 0.125 0.15 0.175 n = 5 n = 10 n = 15 n = 20 n = 25 Figure 7.7: Convolution ... customer at a bank is exponentially dis- tributed with mean service time 2 minutes. Let X be the total service time for 10 customers. Estimate the probability that X>2...

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Introduction to Probability - Chapter 1 pps

Introduction to Probability - Chapter 1 pps

... PROBABILITIES 5 5 10 15 20 25 30 35 40 -1 0 -8 -6 -4 -2 2 4 6 8 10 Figure 1.1: Peter’s winnings in 40 plays of heads or tails. One can understand this calculation as follows: The probability that ... more appropriate to assign a distribution function which assigns probability .513 to the outcome boy and probability .487 to the outcome girl than to assign pro...

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Introduction to Probability - Chapter 2 docx

Introduction to Probability - Chapter 2 docx

... with careful planning one would have to be extremely lucky to be able to stop so cleverly. The second author likes to trace his interest in probability theory to the Chicago World’s Fair of 1933 ... referring to a continuous random variable X (say with a uniform density function), it is customary to say that “X is uniformly distributed on the interval [a, b].” It is also custo...

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Introduction to Probability - Chapter 3 ppt

Introduction to Probability - Chapter 3 ppt

... =  1234 2143  , indicating that a 1 went to a 2 , a 2 to a 1 , a 3 to a 4 , and a 4 to a 3 . Ifwealwayschoosethetoprowtobe1234then, to prescribe the permutation, we need only give the bottom row, with the understanding ... Birthday problem. We now turn to the topic of permutations. Permutations Definition 3.1 Let A be any finite set. A permutation of A is a one -to- one mapping of...

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Introduction to Probability - Chapter 4 doc

Introduction to Probability - Chapter 4 doc

... ✷ 146 CHAPTER 4. CONDITIONAL PROBABILITY Number having The results Disease this disease ++ +– –+ –– d 1 3215 2 110 301 704 100 d 2 2125 396 132 1187 410 d 3 4660 510 3568 73 509 Total 100 00 Table ... drawn is red? 4.3. PARADOXES 175 13 Write a program to allow you to compare the strategies play-the-winner and play-the-best-machine for the two-armed bandit problem of Example...

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Introduction to Probability - Chapter 5 docx

Introduction to Probability - Chapter 5 docx

... result of 100 Bernoulli trials with p =1 /100 0. The expected value of S 100 is λ = 100 (1 /100 0) = .1. The exact probability that S 100 = j is b (100 , 1 /100 0,j), and the Poisson approximation is e −.1 (.1) j j! . In ... distribution to X? If it is appropriate, what probability should we assign to each outcome? (a) Take the first 100 students who enter the cafeteria to eat...

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Introduction to Probability - Chapter 6 doc

Introduction to Probability - Chapter 6 doc

... 61 62 63 64 1 /10 2 /10 4 /10 2 /10 1 /10  . (a) Find E(F ) and V (F ). (b) Define T = F − 62. Find E(T ) and V (T ), and compare these answers with those in part (a). (c) It is decided to report the ... real-valued random vari- able. One way to determine the expected value of φ(X) is to first determine the distribution function of this random variable, and then use the definition of...

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Introduction to Probability - Chapter 8 docx

Introduction to Probability - Chapter 8 docx

... ≤ .21 n(.1) 2 = 21 n . Thus, if n = 100 , P (|A 100 − .3|≥.1) ≤ .21 , or if n = 100 0, P (|A 100 0 − .3|≥.1) ≤ .021 . These can be rewritten as P (.2 <A 100 <.4) ≥ .79 , P (.2 <A 100 0 <.4) ≥ .979 . These ... x) to compute the exact probability that you estimated in Exercise 1. Compare the two results. 3 Write a program to toss a coin 10, 000 times. Let S n be the n...

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Introduction to Probability - Chapter 9 pps

Introduction to Probability - Chapter 9 pps

... want to estimate the probability P (S 1700 > 106 0) = P (S 1700 ≥ 106 1) = P  S ∗ 1700 ≥ 106 0.5 102 0 20  = P (S ∗ 1700 ≥ 2.025) . From Table 9.4, if we interpolate, we would estimate this probability ... or greater. Estimate the probability that, for this to happen, (a) more than 210 tosses are required. (b) less than 190 tosses are required. (c) between 180 and 210 toss...

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