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056Independent Events Cannot Be DisjointSuppose P(A) = 1/3 and P(B) = 1/2. Can A and B be simultaneously independent and mutually exclusive? Prove your answer.概率简单derivation未尝试免费057Independence Is Not TransitiveLet \Omega = \ 1,2,3,4\ with uniform probability. Define A = \ 1,2\ , B = \ 1,3\ , C = \ 3,4\ . (a) Verify that A and B are independent. (b) Verify that B and C are independent. (c) Are A and C independent? What does this say about transitivity of independence?概率中等derivation未尝试免费058XOR Pairwise Independence Without Mutual IndependenceLet X and Y be independent Bernoulli (1/2) random variables. Define Z = X \oplus Y (where \oplus denotes addition modulo 2). (a) Show that Z \sim Bernoulli (1/2). (b) Show that any two of \ X, Y, Z\ are independent. (c) Are X, Y, Z mutually independent? Exhibit a specific joint event that violates the mutual independence condition.概率中等derivation未尝试免费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061Complements of Independent Events Are IndependentLet A and B be independent events. Prove that A c and B c are also independent, i.e., P(A c \cap B c) = P(A c) P(B c).概率简单derivation未尝试免费062Triple Product Can Hold Without Pairwise IndependenceLet \Omega = \ 1,2,\ldots,8\ with uniform probability. Define events: A = \ 1,2,3,4\ , \quad B = \ 1,2,3,5\ , \quad C = \ 1,4,6,7\ . (a) Show that P(A \cap B \cap C) = P(A)P(B)P(C). (b) Check whether each pair (A,B), (A,C), (B,C) is independent. (c) What does this reveal about the relationship between the triple-product condition and pairwise independence?概率简单derivation未尝试免费063A Function of Independent Variables Need Not Be Independent of Its InputsLet X and Y be independent Bernoulli (1/2) random variables. Define W = \max(X, Y). (a) Compute the distribution of W. (b) Determine whether X and W are independent by checking all four joint probabilities P(X = x, W = w) for x, w \in \ 0, 1\ .概率中等derivation未尝试免费064Independence Is Not Closed Under UnionsLet A, B, C be events with A \perp\!\!\perp B and A \perp\!\!\perp C. (a) Prove that if A, B, C are mutually independent, then A \perp\!\!\perp (B \cup C). (b) Now let \Omega = \ 1,2,3,4\ with uniform probability, A = \ 1,2\ , B = \ 1,3\ , C = \ 1,4\ . Verify that A \perp\!\!\perp B and A \perp\!\!\perp C. (c) Show that A and B \cup C are not independent. Why doesn't pairwise independence suffice?概率中等derivation未尝试免费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066Zero-Probability Events Are Independent of EverythingLet (\Omega, \mathcal F , P) be a probability space and let A be an event with P(A) = 0. (a) Prove that A is independent of every event B \in \mathcal F . (b) Let \Omega = \ 1,2,3,4,5,6\ with uniform probability. Set A = \emptyset and B = \ 1,2,3\ . Verify the independence condition directly. (c) Does the same conclusion hold when P(A) = 1? Prove or give a counterexample.概率简单derivation未尝试免费067Identical Marginals Do Not Imply IndependenceThree cards are labeled (0,0), (0,1), and (1,0). One card is drawn uniformly at random. Let X be the first number and Y the second number on the drawn card. (a) Compute P(X = 0), P(X = 1), P(Y = 0), and P(Y = 1). (b) Are the marginal distributions of X and Y the same? (c) Are X and Y independent? Verify by checking whether P(X = x, Y = y) = P(X = x) P(Y = y) holds for all (x, y) \in \ 0,1\ 2.概率中等derivation未尝试免费068Independence Is Not Hereditary to Sub-EventsLet \Omega = \ 1,2,3,4\ with uniform probability. Define A = \ 1,2\ and B = \ 1,3\ . (a) Verify that A and B are independent. (b) Let A 1 = \ 1\ \subseteq A. Determine whether A 1 and B are independent. (c) State a general principle: if A \perp\!\!\perp B and A 1 \subseteq A, does A 1 \perp\!\!\perp B necessarily hold?概率中等derivation未尝试免费069An Event Independent of Itself Must Be TrivialLet A be an event in a probability space. (a) Write the independence condition P(A \cap A) = P(A) P(A) and deduce which values of P(A) satisfy it. (b) On \Omega = \ 1,2,3,4\ with uniform probability, verify your answer by testing A = \ 1\ , A = \ 1,2\ , A = \emptyset, and A = \Omega. (c) Interpret: what does it mean probabilistically for an event to be independent of itself?概率中等derivation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071Independence Is Fragile Under Sample-Space PerturbationA four-card deck contains A\spadesuit, A\heartsuit, K\spadesuit, K\heartsuit. One card is drawn uniformly at random. Let R = 「the card is an Ace」 and S = 「the card is a Spade.」 (a) Compute P(R), P(S), P(R \cap S) and verify that R and S are independent. (b) Now remove K\heartsuit from the deck, leaving a three-card deck \ A\spadesuit, A\heartsuit, K\spadesuit\ with uniform draw. Recompute P(R), P(S), P(R \cap S) and determine whether R and S are still independent. (c) Explain intuitively why removing a single card destroyed independence.概率简单derivation未尝试免费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