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hw3_sol - ECE 139 Part A Probability and Statistics HW#3...

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ECE 139 Probability and Statistics HW#3 - Solutions Part A: ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 1.6.3 P[A] = 1/4, P[B] = 1/8, P[C] = 5/8, P[D] = 3/8, A & B are disjoint. C& D are independent. (a) A & B are disjoint. A B = P[A B] = 0 (i.e. A is a subset of B c .) P[A B] = P[A] + P[B] = 3/8 P[A B c ] = P[A] = 1/4 P[A B c ] = P[B c ] = 1 – P[B] = 1 – 1/8 = 7/8 (b) P[A]×P[B] = 1/4 × 1/8 = 1/32 ≠ 0 = P[A B] A & B are not independent. (c) C & D are independent. P[C D] = P[C]×P[D] = 5/8 × 3/8 = 15/64 P[C D c ] = P[C] - P[C D] = 5/8 – 15/64 = 25/64 → P[C c D c ] = P[(C D) c ] = 1 – P[C D] = 1 – (P[C] + P[D] – P[C D]) = 1 – 5/8 – 3/8 + 15/64 = 15/64 (d) P[C c ]×P[D c ] = (1 – P[C])×(1 – P[D]) = (1 – 5/8) × (1 – 3/8) = 15/64 = P[C c D c ] → Therefore, C c & D c are independent. 1.6.5 S = {1, 2, 3, 4}, P[{1}] = P[{2}] = P[{3}] = P[{4}] = 1/4 Consider: A 1 = {1, 2}, A 2 = {2, 3}, A 3 = {1, 3} P[ A 1 ] = P[A 2 ] = P[A 3 ] = 1/2 P[A 1 A 2 ] = P[{2}] = 1/4 = P[A 1 ]×P[A 2 ] A 1 & A 2 are independent. P[A 1 A 3 ] = P[{1}] = 1/4 = P[A 1 ]×P[A 3 ] A 1 & A 3 are independent.
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