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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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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i 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
hmm, auc = a + c - ac aub = a + b - ab bc = b (c|b) hmm
ya i found P(ab) to be .45 and P(AC) to be .3

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Other answers:

auc = a + c - ac = .7 aub = a + b - ab = .55 and a=b=c=.5 auc = .5 + .5 - ac = .7 aub = .5 + .5 - ab = .55
using the first two you said
ac = .3 ab = .45
bc = .5 (c|b) = .3 c|b = 3/5
i 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)
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1+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
correct im following that
im thinking if we knew c|a: p(ac) = a (c|a)
oh remember P(ABC)=2P(ABC^complement))
right, that most likely allows us to determine the universal set as a whole
.3 = .5 (c|a) .6 = c|a 4+5 = 60% of a as well; .3
ab = a (b|a) .45 = .5 (b|a) .9 = b|a such that 2+5 = .9*.5 = .45
might just have to wolf it to be certain :)
wolfram alpha?
yeah, but i dont know the syntax to write it up
ah ya me neither i dident know it could do probabilitys
there are those on here which can do these in their sleep, im just not one of them :)
haha alright i understand i cant either
using 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)
see what i though was maybe you can do something like: P(AUB)UP(C)?
so P(AUBUC)=P(AUB)+P(C)-(P(AB)UP(AC)) P(AUBUC)=.55+.5...?
im not sure what that last term would be though
yeah, im coming up empty on it as well
here is what i have so far p(A)=P(B)=B(C)=.5 p(AB)=.45 p(AC)=.2 P(BC)=.3
does this look right?
for P(AC) im getting .3
yeah my mistake
but 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
somehow we need two expressions that contain p(ABC) that we know everything except that one
well my professor said that this might be helpful: P(AUBUC) = P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC)
so 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?
yeah we get that for sure it is \(.45\)
i mean it is \(.45-x\)
putting \(x=P(ABC)\)
now we need some other expression involving \(x\)
see i was thinking if you treat AUB = y then could you find P(yUC)? which would be P(AUBUC)
i am thinking some demorgan law
maybe finding \[P(B^cC^c)\]?
am i being stupid, or does it say \(x=2(1-x)\)?
if \(P(ABC)=x\), then \(P((ABC)^c)=1-x\)
oh no
i guess i read it incorrectly
its \[P(ABC) = 2(P(ABC^c))\]
just the complement of C
oooh ok let me look at my diagram again
so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?
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an we know 2x+x=.45
so x=.15. but it should be .30
if i am reading the question correctly (the second time) it is not so bad
why?
because the answer in the back of the book says it should be haha
ok let me look again but i don't see a mistake yet
i have to keep scrolling up it is a pain maybe i will copy here
ya no problem
P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3
.55 = .5 + .5 - .45 so P(AB)=.45
P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)
i get \(3P(ABC)=.45\implies P(ABC)=.15\)
ya 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.
there is a mistake here somewhere
what is P(AC)?
im getting .3 for P(AC)
yeah 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
wait i think i see what is wrong
\[P(ABC)=2P(ABC^c)\] .45=P(AB) .45+P(ABC)+P(ABC^c)
that + should be an = on the bottom one
yeah i got that
so wouldent it be .45/1.5?
oh crap i did it backwards!!
because we will have one P(ABC) and one P(ABC^c) but the equation calls for 2 P(ABC^c)'s ?
lol is that sound logic or am i making stuff up?
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damn sometimes i am dense
no i think it was the other way? ill draw it real quick
really tyro error there, i did it backwards you get \(x=.15, 2x=.3\)
it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!
that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)
AHHHHH lol
got it!
jesus sometimes i think i should lose my math license
damn problem was not that hard only needed P(AB)
haha ya lol wow ok this problem was very frustrating
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it has tripped up everyone i know so far that has tried it and its was a basic multiplication
yeah pretty basic, now that it is finished
lol well hopefully i will be able to recognize this stuff in the future. thank you for your help
yw

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