Please choose the right answers whether it is a,b,c, d, or e. The problem is clear and all details are there. (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) = P(B), or (iii) P(A) = 0, or (iv) P(B) = 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(A n B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we haveP(A n B)P(B)P(A) · P(B)P(A n B)P(A) · P(B)P(A|B) =P(A),or P(BIΑ)-= P(B).%DP(B)P(A)P(A)This finishes the proof.(d) Assume at least one of the the four conditions holds. Condition (i) implies thatP(A n B)P(B)= P(A);Р(An B) 3D Р(А)Р(B).orThis is exactly what independence requires. Similarly, condition (ii) impliesP(B N A)= P(B);orP(B n A) = P(A)P(B).P(A)This is exactly what independence requires. If condition (iii) or (iv) holds then automatically0 < P(A n B) < P(A) = P(A) · P(B) = 0.0 < P(A n B) < P(B) = P(B) · P(A) = 0.orThis 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(A n B)Р(А);P(A n B) = P(A)P(B).orP(B)This is exactly what independence requires. Similarly, condition (ii) impliesP(B n A)= P(B);P(B N A) = P(A)P(B).orP(A)This is exactly what independence requires. If condition (iii) or (iv) holds then automatically0 < P(A n B) < P(A) = P(A) · P(B) = 0.0< ΡPAn B)

Name: Please choose the right answers whether it is a,b,c, d, or e. The prob
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Description: Please choose the right answers whether it is a,b,c, d, or e. The problem is clear and all details are there. (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) = P(B),

(e) is correct. It is given that "two events A,B are independent if and only if at least one of the…

Answered: ts, A, B, are independent if and only…

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ts, 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) = P(B), or (iii) P(A) = 0, or tiv) P(B) = 0.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Concept explainers

Addition Rule of Probability

It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.

Expected Value

When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known as the expectation, is denoted by: E(X).

Probability Distributions

Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosing a card. Each of these events has multiple possibilities. Every such possibility is measured with the help of probability. To be more precise, the probability is used for calculating the occurrence of events that may or may not happen. Probability does not give sure results. Unless the probability of any event is 1, the different outcomes may or may not happen in real life, regardless of how less or how more their probability is.

Basic Probability

The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the probability value 1 indicates that the event will occur. Sum of the probability value must be 1. The probability value is never a negative number. If it happens, then recheck the calculation.

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Please choose the right answers whether it is a,b,c, d, or e. The problem is clear and all details are there.

(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) = P(B), or (iii) P(A) = 0, or (iv) P(B) = 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(A n B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we have
P(A n B)
P(B)
P(A) · P(B)
P(A n B)
P(A) · P(B)
P(A|B) =
P(A),
or P(BIΑ)-
= P(B).
%D
P(B)
P(A)
P(A)
This finishes the proof.
(d) Assume at least one of the the four conditions holds. Condition (i) implies that
P(A n B)
P(B)
= P(A);
Р(An B) 3D Р(А)Р(B).
or
This is exactly what independence requires. Similarly, condition (ii) implies
P(B N A)
= P(B);
or
P(B n A) = P(A)P(B).
P(A)
This is exactly what independence requires. If condition (iii) or (iv) holds then automatically
0 < P(A n B) < P(A) = P(A) · P(B) = 0.
0 < P(A n B) < P(B) = P(B) · P(A) = 0.
or
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(A n B)
Р(А);
P(A n B) = P(A)P(B).
or
P(B)
This is exactly what independence requires. Similarly, condition (ii) implies
P(B n A)
= P(B);
P(B N A) = P(A)P(B).
or
P(A)
This is exactly what independence requires. If condition (iii) or (iv) holds then automatically
0 < P(A n B) < P(A) = P(A) · P(B) = 0.
0< ΡPAn B) <PB)= P(B). P(Α) -0.
or
This is exactly what independence requires. Conversely, if P(A n B) = P(A)P(B), then either (iii) holds or (iv) holds or else the conditional probabilities are defined and we have
P(A n B)
P(B)
P(A n B)
P(A)
ΡA)Ρ(B)
ΡA). Ρ(B)
P(A)
= P(B).
P(A|B) =
= P(A),
or P(B Α) -
%3D
=
P(B)
This finishes the proof.
The correct proof is
(a)
(b)
(c)
(d)
(e)
N/A
(Select One)

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