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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".

Q: Prove that the complement of A and B are independent events?

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Yes.

complement of c

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

.7

(A' âˆ© B') = (A Ãˆ B)'

The complement of a subset B within a set A consists of all elements of A which are not in B.

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

P(A given B')=[P(A)-P(AnB)]/[1-P(B)].

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...).