P(AN B)P(B)= P(A);P(AN B) = P(A)P(B).orThis is exactly what independence requires. Similarly, condition (ii) impliesP(Bn A)= P(B);P(Bn A) = P(A)P(B).orP(A)This is exactly what independence requires. If condition (iii) or (iv) holds then automatically0< 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 thatP(An B)P(B)= P(A);P(AN B) = P(A)P(B).orThis is exactly what independence requires. Similarly, condition (ii) impliesP(BN A)P(B);P(BN A) = P(A)P(B).orP(A)This is exactly what independence requires. If condition (iii) or (iv) holds then automaticallyO< 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) holdsor else the conditional probabilities are defined and we haveP(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 thiscourse.(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 andwe haveР() - 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

Given: 4 conditions given are PA|B=PAPB|A=PBPA=0PB=0

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

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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