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 one year 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))?
 one year 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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Benbburn
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0hmm, auc = a + c  ac aub = a + b  ab bc = b (cb) hmm

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

amistre64
 one year 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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0using the first two you said

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

Benbburn
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1358793002972:dw

amistre64
 one year 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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0correct im following that

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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0oh remember P(ABC)=2P(ABC^complement))

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

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

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

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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0ah ya me neither i dident know it could do probabilitys

amistre64
 one year 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 :)

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0haha alright i understand i cant either

amistre64
 one year 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)

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

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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0im not sure what that last term would be though

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2here is what i have so far p(A)=P(B)=B(C)=.5 p(AB)=.45 p(AC)=.2 P(BC)=.3

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2does this look right?

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0for P(AC) im getting .3

Benbburn
 one year 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2somehow we need two expressions that contain p(ABC) that we know everything except that one

Benbburn
 one year 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)

Benbburn
 one year 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?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2yeah we get that for sure it is \(.45\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2i mean it is \(.45x\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2putting \(x=P(ABC)\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2now we need some other expression involving \(x\)

Benbburn
 one year 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)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2i am thinking some demorgan law

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2am i being stupid, or does it say \(x=2(1x)\)?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2if \(P(ABC)=x\), then \(P((ABC)^c)=1x\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2i guess i read it incorrectly

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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0just the complement of C

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2oooh ok let me look at my diagram again

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2dw:1358826338804:dw

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0so x=.15. but it should be .30

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2if i am reading the question correctly (the second time) it is not so bad

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0because the answer in the back of the book says it should be haha

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2ok let me look again but i don't see a mistake yet

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2i have to keep scrolling up it is a pain maybe i will copy here

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2.55 = .5 + .5  .45 so P(AB)=.45

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2i get \(3P(ABC)=.45\implies P(ABC)=.15\)

Benbburn
 one year 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.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2there is a mistake here somewhere

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0im getting .3 for P(AC)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2yeah 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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0wait i think i see what is wrong

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

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

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0so wouldent it be .45/1.5?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2oh crap i did it backwards!!

Benbburn
 one year 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 ?

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2dw:1358827302867:dw

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2damn sometimes i am dense

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2really tyro error there, i did it backwards you get \(x=.15, 2x=.3\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2jesus sometimes i think i should lose my math license

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2damn problem was not that hard only needed P(AB)

Benbburn
 one year ago
Best ResponseYou've already chosen the best response.0haha ya lol wow ok this problem was very frustrating

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2dw:1358827656432:dw

Benbburn
 one year 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.2yeah pretty basic, now that it is finished

Benbburn
 one year 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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