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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