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".
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.
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...).