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Q: Which equation implies that A and B are independent events?

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Yes.

Given two events, A and B, Pr(A and B) = Pr(A)*Pr(B) if A and B are independent and Pr(A and B) = Pr(A | B)*Pr(B) if they are not.

y=mx+b is the equation for a linear relationship. y= the dependant variable m= the slope of the line x= the independent variable b= the y-intercept

the circles do not overlap at all.

P(A given B)*P(B)=P(A and B), where event A is dependent on event B. Finding the probability of an independent event really depends on the situation (dart throwing, coin flipping, even Schrodinger's cat...).

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Yes.

Given two events, A and B, Pr(A and B) = Pr(A)*Pr(B) if A and B are independent and Pr(A and B) = Pr(A | B)*Pr(B) if they are not.

p(A and B) = p(A) x p(B) for 2 independent events p(A and B and ...N) = p(A) x p(B) x p(C) x ...x p(N) In words, if these are all independent events, find the individual probabilities if each and multiply them all together.

if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof

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y=mx+b is the equation for a linear relationship. y= the dependant variable m= the slope of the line x= the independent variable b= the y-intercept

when the occurance of an event B is not affected by the occurance of event A than we can say that these events are not dependent with each other

the circles do not overlap at all.

Answer: The property that is illustrated is: Symmetric property. Step-by-step explanation: Reflexive property-- The reflexive property states that: a implies b Symmetric Property-- it states that: if a implies b . then b implies a Transitive property-- if a implies b and b implies c then c implies a Distributive Property-- It states that: a(b+c)=ab+ac If HAX implies RIG then RIG implies HAX is a symmetric property.

first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers) P(A' intersect B') = P(B')P(A'|B') by definition = P(B')[1-P(A|B')] since 1 = P(A) + P(A') = P(B')[1 - P(A)] from the first proof * = P(B')P(A') since 1 = P(A) + P(A') conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition. ***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".

P(A given B)*P(B)=P(A and B), where event A is dependent on event B. Finding the probability of an independent event really depends on the situation (dart throwing, coin flipping, even Schrodinger's cat...).

~(A => B) is ~B => ~A That is to say, the converse of "A implies B" is "the converse of B implies the converse of A". In this case: If a shape is not a parallelogram then it is not a rectangle.