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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))?
 one year ago
 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
 one year ago

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BenbburnBest ResponseYou've already chosen the best response.0
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
hmm, auc = a + c  ac aub = a + b  ab bc = b (cb) hmm
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
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
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
bc = .5 (cb) = .3 cb = 3/5
 one year ago

BenbburnBest ResponseYou've already chosen the best response.0
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)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
dw:1358793002972:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
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
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
im thinking if we knew ca: p(ac) = a (ca)
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
right, that most likely allows us to determine the universal set as a whole
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
.3 = .5 (ca) .6 = ca 4+5 = 60% of a as well; .3
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
ab = a (ba) .45 = .5 (ba) .9 = ba such that 2+5 = .9*.5 = .45
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
might just have to wolf it to be certain :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
yeah, but i dont know the syntax to write it up
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
there are those on here which can do these in their sleep, im just not one of them :)
 one year ago

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

amistre64Best ResponseYou've already chosen the best response.0
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)
 one year ago

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

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

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

amistre64Best ResponseYou've already chosen the best response.0
yeah, im coming up empty on it as well
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
here is what i have so far p(A)=P(B)=B(C)=.5 p(AB)=.45 p(AC)=.2 P(BC)=.3
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
does this look right?
 one year ago

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

BenbburnBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
somehow we need two expressions that contain p(ABC) that we know everything except that one
 one year ago

BenbburnBest ResponseYou've already chosen the best response.0
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)
 one year ago

BenbburnBest ResponseYou've already chosen the best response.0
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?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
yeah we get that for sure it is \(.45\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i mean it is \(.45x\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
putting \(x=P(ABC)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
now we need some other expression involving \(x\)
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
i am thinking some demorgan law
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
am i being stupid, or does it say \(x=2(1x)\)?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
if \(P(ABC)=x\), then \(P((ABC)^c)=1x\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i guess i read it incorrectly
 one year ago

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

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

satellite73Best ResponseYou've already chosen the best response.2
oooh ok let me look at my diagram again
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
dw:1358826338804:dw
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
if i am reading the question correctly (the second time) it is not so bad
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
ok let me look again but i don't see a mistake yet
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i have to keep scrolling up it is a pain maybe i will copy here
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
.55 = .5 + .5  .45 so P(AB)=.45
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i get \(3P(ABC)=.45\implies P(ABC)=.15\)
 one year ago

BenbburnBest ResponseYou've already chosen the best response.0
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.
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
there is a mistake here somewhere
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

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

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

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

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

satellite73Best ResponseYou've already chosen the best response.2
oh crap i did it backwards!!
 one year ago

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

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

satellite73Best ResponseYou've already chosen the best response.2
dw:1358827302867:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
damn sometimes i am dense
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
really tyro error there, i did it backwards you get \(x=.15, 2x=.3\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
jesus sometimes i think i should lose my math license
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
damn problem was not that hard only needed P(AB)
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
dw:1358827656432:dw
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.2
yeah pretty basic, now that it is finished
 one year ago

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