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028Factory Posterior After One Good and One Bad ItemA part comes from factory A with prior probability 0.6 and from factory B otherwise. Factory A produces defects with probability 0.1; factory B produces defects with probability 0.4. Two parts from the same unknown factory are inspected and exactly one is defective. What is the posterior probability the parts came from factory A?概率中等数值题未尝试免费032Three-Source Posterior After Alert and Failed ClearanceA transaction comes from source A, B, or C with priors 1/2, 1/3, and 1/6. The alert probabilities are 0.2, 0.4, and 0.8 respectively. Conditional on being alerted, the clearance probabilities are 0.9, 0.6, and 0.25 respectively. Given that the transaction was alerted and then failed clearance, what is the posterior probability it came from source C?概率中等数值题未尝试免费034Repeated Positive Test PosteriorA rare condition has prior probability 1/100. Each test is independent conditional on the true state, with true-positive rate 19/20 and false-positive rate 1/10. If two tests are both positive, what is the posterior probability of the condition?概率困难数值题未尝试面试订阅038Intraday Momentum and Regime ClassificationA quantitative analyst classifies each trading day as either "trending" (probability 0.6) or "mean-reverting" (probability 0.4). - On a trending day, the morning session is positive with probability 0.7, and given a positive morning the afternoon session is also positive with probability 0.8. - On a mean-reverting day, the morning session is positive with probability 0.5, and given a positive morning the afternoon session is positive with probability 0.3. Today both the morning and afternoon sessions are positive. What is the posterior probability that today is a trending day?概率中等数值题未尝试免费040Sequential Signal Updating and the Tower PropertyA quant researcher believes a directional signal has accuracy p that is either 1 3 or 2 3 , each equally likely a priori. On each day the signal independently (given p) predicts the market direction, and is correct with probability p. (a) On Day 1 the signal is correct. What is the posterior P\! (p = \tfrac 2 3 \mid C 1 )? (b) On Day 2 the signal is wrong. Starting from the Day-1 posterior, compute the updated P\! (p = \tfrac 2 3 \mid C 1, W 2 ). (c) Verify the tower property of conditional expectation: show that E[p] = E\! [\,E[p \mid D 1]\, ], where D 1 \in \ C 1, W 1\ denotes the Day-1 outcome. Compute all quantities explicitly.概率困难derivation未尝试面试订阅044Posterior Odds UpdateThe prior probability of a hypothesis is 3/10. An observed signal has likelihood ratio (4/5)/(2/5). What are the posterior odds in favor of the hypothesis?概率中等数值题未尝试免费050Noisy Signal Detection and Evidence ThresholdA hidden signal S is equally likely to be +1 or -1. At each time step you receive a noisy reading: if S = +1 the reading is +1 with probability 2 3 and -1 with probability 1 3 ; if S = -1 the reading is -1 with probability 2 3 and +1 with probability 1 3 . Readings are conditionally independent given S. (a) You observe the sequence (+1, +1, -1). Find the posterior probability P(S = +1 \mid observations ). (b) Starting from the uniform prior, what is the minimum number n of consecutive +1 readings required so that P(S = +1 \mid n consecutive +1) > 0.95?概率困难数值题未尝试免费070Pairwise and Triple Independence Without Mutual Independence of Four EventsLet \Omega consist of all binary strings of length 4 with an even number of 1s, each equally likely: \Omega = \ 0000,\, 0011,\, 0101,\, 0110,\, 1001,\, 1010,\, 1100,\, 1111\ . Define events A i = \ \omega \in \Omega : \omega i = 1\ for i = 1,2,3,4. (a) Show that P(A i) = 1/2 for each i. (b) Verify that every pair is independent: P(A i \cap A j) = 1/4 for all i j. (c) Verify that every triple is independent: P(A i \cap A j \cap A k) = 1/8 for all distinct i,j,k. (d) Compute P(A 1 \cap A 2 \cap A 3 \cap A 4) and compare with P(A 1) P(A 2) P(A 3) P(A 4). Are the four events mutually independent?概率困难derivation未尝试面试订阅078Simpson's Paradox in a Clinical TrialA clinical trial tests a drug on two subgroups. In subgroup A (mild cases): 81/87 (93%) of treated patients recover vs 234/270 (87%) of untreated. In subgroup B (severe cases): 192/263 (73%) of treated patients recover vs 55/80 (69%) of untreated. The drug improves recovery rates in both subgroups. Now compute the overall recovery rates (combining subgroups) for treated vs untreated. Explain the apparent contradiction and identify the lurking variable that causes it.概率中等derivation未尝试免费085The Inspection Paradox (Bus Waiting Time)Buses arrive at a stop according to a Poisson process with rate (so inter-arrival times are iid Exp ( ) with mean 1/ ). You arrive at the bus stop at a uniformly random time, independent of the bus schedule. Let L be the length of the inter-arrival interval that contains your arrival time — i.e., the time between the last bus before you arrived and the next bus after. (a) Find E[L]. Explain why it is **not** 1/ despite inter-arrival times having mean 1/ . (b) Find the expected waiting time E[W] until the next bus, where W is the time from your arrival until the next bus. (c) A city official surveys bus riders and asks how long they waited. If the reported average is 1/ , should the transit authority be surprised? Explain using the inspection paradox.概率困难derivation未尝试面试订阅090Borel's Paradox: Conditioning on Measure-Zero EventsLet (\Theta, \Phi) be uniformly distributed on the unit sphere S 2, where \Theta \in [0, 2 ) is the longitude and \Phi \in [0, ] is the colatitude, with joint density f( , \phi) = 1 4 \sin \phi. (a) Compute the conditional distribution of \Phi given \Theta = 0 by treating \Theta as the conditioning variable (i.e., compute f(\phi \mid = 0)). (b) Now reparameterize: let X = \cos(\Theta) \sin(\Phi), Y = \sin(\Theta) \sin(\Phi), Z = \cos(\Phi). The great circle \ \Theta = 0\ can equivalently be described as \ Y = 0, X 0\ . Compute the conditional distribution of \Phi given Y = 0 and X > 0. Is it the same as in part (a)? (c) Explain why the two answers differ. What does this tell us about the meaning of 'conditioning on a measure-zero event,' and what is the correct mathematical framework for resolving this ambiguity?概率困难derivation未尝试免费105Expected Number of Suits in a Poker HandA 5-card hand is dealt from a standard 52-card deck. Let S be the number of distinct suits represented in the hand. Using indicator random variables, find E[S].概率困难derivation未尝试面试订阅154Expected Number of Birthday-Collision PairsIn a group of n people whose birthdays are independent and uniform on \ 1,\ldots,365\ , let X be the number of unordered pairs (i,j) with i < j who share a birthday. Using indicator random variables, find E[X]. Then determine the smallest n for which E[X] \ge 1.概率中等derivation未尝试免费155Variance of Birthday-Collision Pair CountContinuing from the setup of the expected collision-pair count: n people have independent uniform birthdays on \ 1,\ldots,d\ . Define X = \sum i<j 1 [B i = B j]. (a) Compute Var (X). (b) A surprising intermediate step: show that Cov ( 1 [B i = B j],\, 1 [B j = B k]) = 0 for distinct i,j,k even though the two indicators share the index j. Explain intuitively why this zero covariance holds. (c) For d = 365 and n = 28, compute Var (X) numerically and give the coefficient of variation \sigma X / E[X].概率困难derivation未尝试面试订阅158Triple Birthday Collision ThresholdIn a room of n people with birthdays uniform on \ 1,\ldots,365\ , let A be the event that at least three people share the same birthday. (a) Using a Poisson approximation (model the occupancy of each day as an independent Poisson (n/365) variable), derive an approximate formula for P(A). (b) Find the smallest n such that P(A) \ge 1 2 under this approximation.概率中等数值题未尝试免费160Expected and Variance of Distinct Birthday CountAmong n people whose birthdays are independent and uniform on \ 1, \ldots, d\ , let D be the number of distinct birthdays observed. (a) Derive E[D] using indicator random variables. (b) Derive Var (D). You will need P( day j and day k both occupied ) for j \ne k. (c) For n = 100 and d = 365, compute E[D], Var (D), and the expected number of "collision people" n - D (people whose birthday coincides with at least one other person). (d) Is E[n - D] the same as the expected number of collision pairs \binom n 2 /d from the indicator-pair approach? Explain the distinction.概率困难derivation未尝试面试订阅164Exactly One Pair Among Four BirthdaysFour people have independent uniform birthdays on a 365-day calendar. What is the probability that there is exactly one matching pair and no larger collision?概率困难derivation未尝试面试订阅169Expected Pairs Within Two DaysOn a 365-day circular calendar, what is the expected number of unordered pairs among n independent uniform birthdays whose birthdays are at circular distance at most 2 days?概率困难derivation未尝试面试订阅170Collision Risk for the Next Arrival Given Distinct Existing BirthdaysSuppose n existing birthdays are all distinct on a 365-day calendar. A new person's birthday is then drawn uniformly and independently. What is the probability the new arrival creates an exact collision?概率困难derivation未尝试面试订阅194Variance of the Number of Occupied UrnsFour distinguishable balls are thrown independently and uniformly at random into 3 distinguishable urns. Let N be the number of nonempty urns. Find Var (N). Give an exact fraction.概率困难数值题未尝试免费