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anonymous
 4 years ago
Can anyone provide me with some guidance on this problem? What is P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3, and P(ABC)=2P(ABC^(complement))?
anonymous
 4 years ago
Can anyone provide me with some guidance on this problem? What is P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3, and P(ABC)=2P(ABC^(complement))?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have tried calculating some of the other things as well but I dont think they got me any closer to an answer. I also know the answer should be .3 but i dont understand how to get to that

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0hmm, auc = a + c  ac aub = a + b  ab bc = b (cb) hmm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya i found P(ab) to be .45 and P(AC) to be .3

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0auc = a + c  ac = .7 aub = a + b  ab = .55 and a=b=c=.5 auc = .5 + .5  ac = .7 aub = .5 + .5  ab = .55

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0using the first two you said

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0bc = .5 (cb) = .3 cb = 3/5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i also know my professor said that this may be useful: P(AUBUC) = P(A)+P(B)+P(C)P(AB)P(AC)P(BC)+P(ABC)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358793002972:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.01+2+4+5 = .5 2+3+5+6 = .5 4+5+6+7 = .5 1+2+4+5+6+7 = .7 1+2+3+4+5+6 = .55 5+6 = 3/5 of .5 = .3 and the idea is to determine 5 is what ive got rolling around in me head

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0correct im following that

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0im thinking if we knew ca: p(ac) = a (ca)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh remember P(ABC)=2P(ABC^complement))

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0right, that most likely allows us to determine the universal set as a whole

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0.3 = .5 (ca) .6 = ca 4+5 = 60% of a as well; .3

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0ab = a (ba) .45 = .5 (ba) .9 = ba such that 2+5 = .9*.5 = .45

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0might just have to wolf it to be certain :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, but i dont know the syntax to write it up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah ya me neither i dident know it could do probabilitys

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0there are those on here which can do these in their sleep, im just not one of them :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha alright i understand i cant either

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0using your teacher hint P(AUBUC) = P(A)+P(B)+P(C)P(AB)P(AC)P(BC)+P(ABC) P(AUBUC)P(A)P(B)P(C)+P(AB)+P(AC)+P(BC) = P(ABC) we know all those except for P(aubuc)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see what i though was maybe you can do something like: P(AUB)UP(C)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so P(AUBUC)=P(AUB)+P(C)(P(AB)UP(AC)) P(AUBUC)=.55+.5...?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im not sure what that last term would be though

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, im coming up empty on it as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here is what i have so far p(A)=P(B)=B(C)=.5 p(AB)=.45 p(AC)=.2 P(BC)=.3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does this look right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for P(AC) im getting .3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but at this point im not sure what to do. i have computed a few other things but im not completly confident if i did them legitimate ways or not lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0somehow we need two expressions that contain p(ABC) that we know everything except that one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well my professor said that this might be helpful: P(AUBUC) = P(A)+P(B)+P(C)P(AB)P(AC)P(BC)+P(ABC)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i was trying to come up with a way to compute P(AUBUC) and thought maybe it was something along the lines of P((AUB)UC) or somthing?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah we get that for sure it is \(.45\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i mean it is \(.45x\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now we need some other expression involving \(x\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see i was thinking if you treat AUB = y then could you find P(yUC)? which would be P(AUBUC)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am thinking some demorgan law

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe finding \[P(B^cC^c)\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0am i being stupid, or does it say \(x=2(1x)\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if \(P(ABC)=x\), then \(P((ABC)^c)=1x\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess i read it incorrectly

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its \[P(ABC) = 2(P(ABC^c))\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just the complement of C

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oooh ok let me look at my diagram again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358826338804:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so x=.15. but it should be .30

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if i am reading the question correctly (the second time) it is not so bad

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because the answer in the back of the book says it should be haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok let me look again but i don't see a mistake yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have to keep scrolling up it is a pain maybe i will copy here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0.55 = .5 + .5  .45 so P(AB)=.45

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get \(3P(ABC)=.45\implies P(ABC)=.15\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya ok i can see that for sure and i follow the logic as well. so im gunna go with that and email him about it to see what he says.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is a mistake here somewhere

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im getting .3 for P(AC)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i must be reading it incorrectly because if p(ABC^c)=.3 and P(ABC)=.15 so that P(AB) = .45 then P(A) is too large

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait i think i see what is wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[P(ABC)=2P(ABC^c)\] .45=P(AB) .45+P(ABC)+P(ABC^c)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that + should be an = on the bottom one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so wouldent it be .45/1.5?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh crap i did it backwards!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because we will have one P(ABC) and one P(ABC^c) but the equation calls for 2 P(ABC^c)'s ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol is that sound logic or am i making stuff up?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358827302867:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0damn sometimes i am dense

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i think it was the other way? ill draw it real quick

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0really tyro error there, i did it backwards you get \(x=.15, 2x=.3\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0jesus sometimes i think i should lose my math license

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0damn problem was not that hard only needed P(AB)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha ya lol wow ok this problem was very frustrating

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358827656432:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it has tripped up everyone i know so far that has tried it and its was a basic multiplication

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah pretty basic, now that it is finished

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol well hopefully i will be able to recognize this stuff in the future. thank you for your help
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