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P(A); P(AN B) = P(A)P(B). or P(B) This is exactly what independence requires. Similarly, condition (ii) implies P(BN A) P(A) P(B); or P(BN A) = P(A)P(B). This is exactly what independence requires. If condition (iii) or (iv) holds then automatically 0< ΡPAΠ Β) < P(4) P(4) . P(B)-0 or 0< PAn Β) < PB) - PB) . PA) =0. This is exactly what independence requires and the proof is finished. (e) Assume at least one of the the four conditions holds. Condition (i) implies that P(An B) = P(A); P(AN B) = P(A)P(B). or

P(A); P(AN B) = P(A)P(B). or P(B) This is exactly what independence requires. Similarly, condition (ii) implies P(BN A) P(A) P(B); or P(BN A) = P(A)P(B). This is exactly what independence requires. If condition (iii) or (iv) holds then automatically 0< ΡPAΠ Β) < P(4) P(4) . P(B)-0 or 0< PAn Β) < PB) - PB) . PA) =0. This is exactly what independence requires and the proof is finished. (e) Assume at least one of the the four conditions holds. Condition (i) implies that P(An B) = P(A); P(AN B) = P(A)P(B). or

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter2: Matrices
Section2.5: Markov Chain
Problem 16E: Consumer Preference In a population of 100,000 consumers, there are 20,000 users of Brand A, 30,000...
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P(AN B) P(B) = P(A); P(AN B) = P(A)P(B). or This is exactly what independence requires. Similarly, condition (ii) implies P(Bn A) = P(B); P(Bn A) = P(A)P(B). or P(A) This is exactly what independence requires. If condition (iii) or (iv) holds then automatically 0< PAn B)< PA)P(Α) - P(B) -0 or 0< P(Α Ω Β) < P(B) P(B).P(Α) 0. This is exactly what independence requires and the proof is finished. (e) Assume at least one of the the four conditions holds. Condition (i) implies that P(An B) P(B) = P(A); P(AN B) = P(A)P(B). or This is exactly what independence requires. Similarly, condition (ii) implies P(BN A) P(B); P(BN A) = P(A)P(B). or P(A) This is exactly what independence requires. If condition (iii) or (iv) holds then automatically O< P(ANB) < P(A) = P(A) · P(B) = 0. or 0< P(An B) < P(B) = P(B) · P(A) = 0. This is exactly what independence requires. Conversely, if P(An B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we have P(A) · P(B) P(A) · P(B) P(A) P(An B) P(An B) P(A|B) = P(B) P(B) = P(A), or P(B|A) = P(A) - Р(В). This finishes the proof. The correct proof is (a) (b) (c) (d) (e) N/A (Select One)
Transcribed Image Text:P(AN B) P(B) = P(A); P(AN B) = P(A)P(B). or This is exactly what independence requires. Similarly, condition (ii) implies P(Bn A) = P(B); P(Bn A) = P(A)P(B). or P(A) This is exactly what independence requires. If condition (iii) or (iv) holds then automatically 0< PAn B)< PA)P(Α) - P(B) -0 or 0< P(Α Ω Β) < P(B) P(B).P(Α) 0. This is exactly what independence requires and the proof is finished. (e) Assume at least one of the the four conditions holds. Condition (i) implies that P(An B) P(B) = P(A); P(AN B) = P(A)P(B). or This is exactly what independence requires. Similarly, condition (ii) implies P(BN A) P(B); P(BN A) = P(A)P(B). or P(A) This is exactly what independence requires. If condition (iii) or (iv) holds then automatically O< P(ANB) < P(A) = P(A) · P(B) = 0. or 0< P(An B) < P(B) = P(B) · P(A) = 0. This is exactly what independence requires. Conversely, if P(An B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we have P(A) · P(B) P(A) · P(B) P(A) P(An B) P(An B) P(A|B) = P(B) P(B) = P(A), or P(B|A) = P(A) - Р(В). This finishes the proof. The correct proof is (a) (b) (c) (d) (e) N/A (Select One)
(2) Show that two events, A, B, are independent if and only if at least one of the following holds: (i) P(A|B) = P(A) or (ii) P(B|A) — Р(B), or (i) P(А) — 0, or (iv) P(в) — 0. The following proofs are proposed. (a) Although the result is true, to prove it, it requires deeper results from Calculus that are beyond the prerequisite of this course. (b) The statement of the problem is, in fact, false. (c) If P(AN B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we have Р() - Р(В) P(AN B) Р(B) P(A) · P(B) Р(В) P(AN B) P(A) P(A|B) = P(A), or P(B|A) = = P(B). P(A) This finishes the proof. (d) Assume at least one of the the four conditions holds. Condition (i) implies that
Transcribed Image Text:(2) Show that two events, A, B, are independent if and only if at least one of the following holds: (i) P(A|B) = P(A) or (ii) P(B|A) — Р(B), or (i) P(А) — 0, or (iv) P(в) — 0. The following proofs are proposed. (a) Although the result is true, to prove it, it requires deeper results from Calculus that are beyond the prerequisite of this course. (b) The statement of the problem is, in fact, false. (c) If P(AN B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we have Р() - Р(В) P(AN B) Р(B) P(A) · P(B) Р(В) P(AN B) P(A) P(A|B) = P(A), or P(B|A) = = P(B). P(A) This finishes the proof. (d) Assume at least one of the the four conditions holds. Condition (i) implies that
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