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026Conflicting Two-Screen UpdateA loan is distressed with prior probability 0.3. Screen A flags a distressed loan with probability 0.8 and flags a healthy loan with probability 0.1. Any flagged loan then goes to screen B, which passes a distressed loan with probability 0.25 and passes a healthy loan with probability 0.70. Given that A flagged the loan and B did not pass it, what is the posterior probability the loan is distressed?概率简单数值题未尝试免费035Positive Then Negative PosteriorA model-risk event has prior probability 1/50. Conditional on the event, the probabilities of a positive screen and then a negative manual review are 9/10 and 1/10. Conditional on no event, those probabilities are 1/5 and 4/5. If the observed sequence is positive then negative, what is the posterior event probability?概率困难数值题未尝试面试订阅037Two Correct Calls PosteriorAn analyst is high-skill with prior probability 2/5. A high-skill analyst makes a correct call with probability 4/5; a low-skill analyst does so with probability 11/20. If two independent calls are both correct, what is the posterior probability the analyst is high-skill?概率简单数值题未尝试免费039Double Alert PosteriorA transaction is malicious with prior probability 1/40. Two independent alert engines each fire with probability 7/10 on malicious transactions and 1/20 on benign ones. If both engines fire, what is the posterior malicious probability?概率中等数值题未尝试免费045Multi-Analyst Signal Fusion and Sequential UpdatingA trading desk receives directional predictions from three independent analysts. The market will either go "up" or "down", each with prior probability 1 2 . Each analyst independently (given the true state) predicts the correct direction with the following probabilities: - Analyst 1: accuracy 0.8 - Analyst 2: accuracy 0.7 - Analyst 3: accuracy 0.9 (a) All three analysts predict "up". What is P( up \mid all three say up )? (b) Analysts 1 and 2 predict "up", but Analyst 3 predicts "down". What is P( up \mid U 1, U 2, D 3)? (c) Show that the posterior in part (b) can be computed by sequential Bayesian updating — updating on one analyst's signal at a time — and that the final answer does not depend on the order of updates. Demonstrate this by computing the posterior via two different orderings.概率困难derivation未尝试面试订阅046Two Red Draws PosteriorA hidden urn is equally likely to be A or B. Urn A gives red with probability 4/5 and urn B gives red with probability 1/3. Two independent draws with replacement are both red. What is the posterior probability the urn is A?概率简单数值题未尝试免费049Mystery Coin IdentificationA box contains three coins, each equally likely to be selected: - **Coin F** (fair): P(H) = 1 2 - **Coin H** (heads-biased): P(H) = 3 4 - **Coin T** (tails-biased): P(H) = 1 4 You pick one coin uniformly at random and flip it three times, obtaining the sequence H, T, H (flips are conditionally independent given the coin). (a) Find the posterior probability that the coin is each type. (b) What is the conditional probability that the fourth flip is heads?概率困难数值题未尝试免费055Independent Events Become Dependent Under ConditioningLet A and B be independent events with P(A) = P(B) = 1/2. Define D = A \triangle B (exactly one of A, B occurs). (a) Compute P(D), P(A \mid D), and P(A \cap B \mid D). (b) Are A and B conditionally independent given D? (c) Are A and B conditionally independent given D c?概率困难derivation未尝试面试订阅059All Triples Independent but Quadruple NotLet \Omega = \ 0, 1, 2, \ldots, 7\ with uniform probability P(\ \omega\ ) = 1/8. Write each \omega in binary as (b 2, b 1, b 0). Define events: A = \ \omega : b 0 = 1\ , \quad B = \ \omega : b 1 = 1\ , \quad C = \ \omega : b 2 = 1\ , \quad D = \ \omega : b 0 \oplus b 1 \oplus b 2 = 1\ . (a) Show that A, B, C are mutually independent. (b) Show that any triple chosen from \ A,B,C,D\ is mutually independent. (c) Show that A, B, C, D are not mutually independent.概率困难derivation未尝试面试订阅060Independence from Boolean Combinations Requires Mutual IndependenceLet A, B, C be events. (a) Prove that if A, B, C are mutually independent, then A is independent of B \cap C c. (b) Now let \Omega = \ 1,2,3,4\ with uniform probability, A = \ 1,2\ , B = \ 1,3\ , C = \ 1,4\ . Verify that A, B, C are pairwise independent but not mutually independent. (c) Compute P(A \cap (B \cap C c)) and P(A) P(B \cap C c). Does the independence A \perp\!\!\perp (B \cap C c) hold?概率困难derivation未尝试面试订阅065Mixtures of Independent Distributions Destroy IndependenceA biased coin lands heads with probability 1/2. If heads, set (X, Y) = (1, 1). If tails, draw X and Y independently, each Bernoulli (1/2). (a) Compute the full joint distribution of (X, Y). (b) Show that X and Y have the same marginal distribution. (c) Are X and Y independent? Prove your answer. (d) Compute P(X = 1 \mid Y = 0) and compare it to P(X = 1). Interpret the result.概率困难derivation未尝试面试订阅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077The Tuesday Boy ProblemA family has exactly two children. You learn that at least one child is a boy who was born on a Tuesday. Assume each child is equally likely to be a boy or a girl, and equally likely to be born on any of the seven days of the week, all independently. What is the probability that both children are boys?概率简单数值题未尝试免费079Four-Door Monty HallThere are four doors: one hides a car, the other three hide goats. You pick Door 1. The host, who knows where the car is, opens one of the remaining doors to reveal a goat (choosing uniformly at random among the goat doors he can open). He opens Door 4. You are now given two options: (a) stick with Door 1, or (b) switch to one of the two unopened doors (Door 2 or Door 3), chosen uniformly at random. What is the probability of winning the car under each option? Is there a third strategy that does even better?概率中等数值题未尝试免费080The Two-Envelope ParadoxTwo envelopes each contain a positive amount of money; one contains exactly twice the other. You pick one envelope at random and find it contains x dollars. The naive argument says: the other envelope is equally likely to contain 2x or x/2, so the expected value of switching is (1/2)(2x) + (1/2)(x/2) = 5x/4 > x, and you should always switch — but this leads to the absurd conclusion that you should switch back and forth indefinitely. (a) Identify the precise flaw in the naive argument. (b) Suppose the smaller amount S is drawn from a known proper prior distribution with E[S] = < . Show that the unconditional expected gain from switching is zero. (c) Explain why conditional on observing x, it *can* be rational to switch for some values of x and not others.概率困难derivation未尝试面试订阅081Bertrand's Box ParadoxThree boxes each contain two coins. Box 1 has two gold coins, Box 2 has one gold and one silver coin, and Box 3 has two silver coins. You pick a box uniformly at random, then draw one coin at random from that box. The coin you draw is gold. What is the probability that the other coin in the same box is also gold?概率简单数值题未尝试免费082The Sleeping Beauty ProblemSleeping Beauty participates in the following experiment. On Sunday she is put to sleep. A fair coin is flipped. If it lands Heads, she is woken on Monday only. If it lands Tails, she is woken on Monday and again on Tuesday (with her memory of the Monday awakening erased before Tuesday). Each time she wakes, she is asked: 'What is your credence that the coin landed Heads?' She knows the full protocol. (a) Present the argument that her answer should be 1/3 (the 'thirder' position). (b) Present the argument that her answer should be 1/2 (the 'halfer' position). (c) Suppose the experiment is repeated 1000 times independently. If Beauty bets \1 at even odds on Heads each time she is woken, what is her expected net gain or loss over all awakenings? What does this imply about the two positions?概率中等derivation未尝试免费083The Necktie ParadoxTwo players, Alice and Bob, each receive a necktie as a gift. The prices of the two neckties are different positive values. Neither player knows either price. They agree to compare: whoever has the cheaper necktie wins the other's necktie. Alice reasons: 'If my necktie costs x, then I either gain a necktie worth more than x or lose a necktie worth x. Since each is equally likely, my expected gain from the game is positive.' Bob makes the identical argument. Both conclude the game favors them — a contradiction since this is a zero-sum exchange. (a) Identify the precise flaw in Alice's reasoning. (b) Suppose the two necktie prices are drawn by a third party as V and 2V where V \sim Uniform (1, 100), assigned randomly to Alice and Bob. If Alice sees her necktie has price tag x, what is her expected gain from the game as a function of x? Show that the unconditional expected gain is zero.概率中等derivation未尝试免费084Nontransitive DiceThree dice have the following faces. Die A: 2, 2, 4, 4, 9, 9 . Die B: 1, 1, 6, 6, 8, 8 . Die C: 3, 3, 5, 5, 7, 7 . Each die is fair (all six faces equally likely). Two players each choose a die and roll it; the higher number wins. (a) Compute P(A > B), P(B > C), and P(C > A). (b) Show that these dice are nontransitive: A tends to beat B, B tends to beat C, yet C tends to beat A. (c) You are in a game where your opponent chooses a die first, then you choose. Which die should you pick in each case, and what is your winning probability?概率中等数值题未尝试免费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