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072Independent Events Covering the Sample SpaceLet A and B be independent events with P(A \cup B) = 1. (a) Using the independence condition and inclusion-exclusion, prove that (1 - P(A))(1 - P(B)) = 0. (b) What does this force about P(A) and P(B)? (c) On \Omega = \ 1,2,3,4\ with uniform probability, find events A and B that are independent, satisfy P(A \cup B) = 1, and have P(A) = 1 while P(B) = 1/2. Verify all three conditions.概率简单derivation未尝试免费073Divisibility Events and the Containment TrapLet \Omega = \ 0, 1, 2, \ldots, 11\ with uniform probability P(\ k\ ) = 1/12 for each k. Define three events based on divisibility: A = \ k \in \Omega : 2 \mid k\ (even numbers), B = \ k \in \Omega : 3 \mid k\ (multiples of 3), C = \ k \in \Omega : 4 \mid k\ (multiples of 4). (a) List each event explicitly and compute P(A), P(B), P(C). (b) For each of the three pairs (A,B), (A,C), (B,C), determine whether the pair is independent by computing P( intersection ) and comparing with the product of marginals. (c) Identify which pair fails independence and explain the structural reason.概率中等derivation未尝试免费074Parity Bit Breaks Four-Wise IndependenceLet \Omega = \ 0,1,\ldots,15\ with uniform probability, representing all 4-bit binary strings b 1 b 2 b 3 b 4. Define four events: A i = \ \omega \in \Omega : b i(\omega) = 1\ for i = 1,2,3, and A 4 = \ \omega \in \Omega : b 1 \oplus b 2 \oplus b 3 = 1\ where \oplus denotes XOR. (a) Show that each A i has probability 1/2. (b) Prove that \ A 1, A 2, A 3, A 4\ is 3-wise independent: for every subset of size 3, the intersection has probability (1/2) 3. (c) Compute P(A 1 \cap A 2 \cap A 3 \cap A 4) and show that 4-wise independence fails. (d) Explain why the parity event A 4 is fundamentally constrained by A 1, A 2, A 3.概率困难derivation未尝试免费075Fixed Points of a Random Permutation Are Not IndependentA permutation of \ 1,2,3,4\ is chosen uniformly at random (each of the 4! = 24 permutations equally likely). For i = 1,2,3,4, define the event A i = \ (i) = i\ (element i is a fixed point). (a) By counting, show that P(A i) = 1/4 for every i, and P(A i \cap A j) = 1/12 for every pair i j. (b) Are A i and A j independent? (c) Compute P(A i \cap A j \cap A k) for distinct i,j,k and P(A 1 \cap A 2 \cap A 3 \cap A 4). (d) Despite the failure of pairwise independence, verify the classical inclusion-exclusion identity: P\bigl(\bigcup i=1 4 A i\bigr) = 1 - 1 2! + 1 3! - 1 4! .概率困难derivation未尝试免费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097Bertrand's Paradox: The Random Chord ProblemConsider a circle of radius r with an inscribed equilateral triangle. A chord is drawn 'at random.' What is the probability that the chord is longer than the side of the triangle? Compute the answer under each of the following three methods of selecting a random chord: (a) **Random endpoints:** Fix one endpoint on the circle and choose the other endpoint uniformly on the circumference. (b) **Random midpoint:** Choose the midpoint of the chord uniformly in the disk. (c) **Random radius:** Choose a radius, then choose a point uniformly along that radius as the midpoint of the chord. For each method, set up and evaluate the relevant integral or geometric argument.概率简单derivation未尝试免费102First Ace vs. First KingA standard 52-card deck is shuffled uniformly at random. What is the probability that the first ace appears before the first king when dealing cards from the top?概率简单数值题未尝试免费108Adjacent Cards in a Shuffled DeckA standard 52-card deck is shuffled uniformly at random and laid out in a row. What is the probability that the ace of spades and the king of spades are adjacent (next to each other)?概率中等数值题未尝试免费109Expected Wait for the First HeartA standard 52-card deck is shuffled uniformly at random. Cards are turned over one at a time from the top. Let X be the position of the first heart. Find E[X].概率中等derivation未尝试免费129Doubles Given an Even SumYou roll two fair six-sided dice and are told that their sum is even. What is the probability that both dice show the same number?概率中等数值题未尝试免费130Expected Distinct Faces on Four DiceYou roll four fair six-sided dice. What is the expected number of distinct face values that appear among the four dice?概率困难数值题未尝试免费134Near-Even Split in Seven Coin FlipsYou flip 7 fair coins. What is the probability that the number of heads and the number of tails differ by at most 1?概率中等数值题未尝试免费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157Non-Uniform Birthdays Increase Collision ProbabilitySuppose d days have birthday probabilities p 1, p 2, \ldots, p d with \sum j p j = 1 (not necessarily uniform). For n people whose birthdays are independent draws from this distribution: (a) Show that for n = 2, P( collision ) = \sum j=1 d p j 2 \ge 1 d , with equality if and only if all p j = 1 d . (b) Deduce that the uniform distribution minimizes the collision probability among all distributions on d days. Give a one-line intuitive explanation for why non-uniformity helps collisions.概率中等derivation未尝试免费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183Collision Probability in the Occupancy ModelFive distinguishable balls are thrown independently and uniformly at random into 12 distinguishable urns. What is the probability that at least two balls land in the same urn? Give an exact fraction.概率中等数值题未尝试免费