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".
Without knowing what A and B are then it is not really possible to know how to prove that they are independent events. The information on what A and B are is needed.
the intolerable acts. no representation in parliament. didnt wanna be involved wit englands wars.
Which events did NOT happen in the 20th century No. A The War of 1812 = 1812, 19th century No. B The Spanish-American War = 1898, 19th century YES. C The Great Depression = 1930s, 20th century No. D invention of the first Kodak camera = first sold 1888, 19th century No. E invention of the cotton gin = patient, 1793, 18th century
George B. McClellan was a Union
Brownell
100
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' ∩ B') = (A È 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.
not (b or c) = (not b) and (not c)