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So .37*.25=.0925 But the answer key for a similar problem( only difference is .47 instead of .4), the answer is .2775. The answers shouldnt differ that much right?
why are you multiplying those two numbers?
isnt that how you get the probability for AND?
not in this case
Then Im completely lost.
your events are NOT independent
I think that's the concept that I dont quite understand
would the formula be P(A U B)/P(A)?
use \[P(A^c\cap B^c)=1-P(A\cup B)\]
so the 1-P(none) formula
for any event E \[P(E)=1-P(E^c)\] or \[P(E^c)=1-P(E)\]
so how exactly does not being independent effect the problem?
events A and B are independent iff \[P(A\cap B)=P(A)P(B)\]
you cant use this formula since A and B are not independent
So in terms of a venn diagram, independent is two circles that dont intersect. and not independent has two circles that do intersect. and in this problem the intersection is .4? So I need to find the sum of each part of the venn diagram, subtract from one (to get the number outside the diagram) which would be the "neither". Correct?
independent events intersect (most of the time)
if A and B are independent and P(A)>0 and P(B)>0 then A and B have to intersect.
use \[P(A^c\cap B^c)=1-P(A\cup B)=1-[P(A)+P(B)-P(A\cap B)]\]
|dw:1321204843404:dw| Is what Im picturing Basically, I was doing P(Ac∩Bc)=1−P(A∪B)=1−[P(A)+P(B)−P(A∩B)] except that mine adds instead of subtracts the intersection.
so to get neither dont you need to get rid of all that is in the circles? including the intersection?
wouldnt that be adding the intersection? so 1−[P(A)+P(B)+P(A∩B)] instead of 1−[P(A)+P(B)−P(A∩B)]
I see nevermind
|dw:1321205205257:dw| So the answer is .02? that's seems really small.(compared to the other answer)
it is .02
I would check your key again...if you just change from .4 to .47 then .2775 is not the answer