anonymous
  • anonymous
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))?
Statistics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
amistre64
  • amistre64
hmm, auc = a + c - ac aub = a + b - ab bc = b (c|b) hmm
anonymous
  • anonymous
ya i found P(ab) to be .45 and P(AC) to be .3

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amistre64
  • amistre64
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
anonymous
  • anonymous
using the first two you said
amistre64
  • amistre64
ac = .3 ab = .45
amistre64
  • amistre64
bc = .5 (c|b) = .3 c|b = 3/5
anonymous
  • anonymous
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)
amistre64
  • amistre64
|dw:1358793002972:dw|
amistre64
  • amistre64
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
anonymous
  • anonymous
correct im following that
amistre64
  • amistre64
im thinking if we knew c|a: p(ac) = a (c|a)
anonymous
  • anonymous
oh remember P(ABC)=2P(ABC^complement))
amistre64
  • amistre64
right, that most likely allows us to determine the universal set as a whole
amistre64
  • amistre64
.3 = .5 (c|a) .6 = c|a 4+5 = 60% of a as well; .3
amistre64
  • amistre64
ab = a (b|a) .45 = .5 (b|a) .9 = b|a such that 2+5 = .9*.5 = .45
amistre64
  • amistre64
might just have to wolf it to be certain :)
anonymous
  • anonymous
wolfram alpha?
amistre64
  • amistre64
yeah, but i dont know the syntax to write it up
anonymous
  • anonymous
ah ya me neither i dident know it could do probabilitys
amistre64
  • amistre64
there are those on here which can do these in their sleep, im just not one of them :)
anonymous
  • anonymous
haha alright i understand i cant either
amistre64
  • amistre64
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)
anonymous
  • anonymous
see what i though was maybe you can do something like: P(AUB)UP(C)?
anonymous
  • anonymous
so P(AUBUC)=P(AUB)+P(C)-(P(AB)UP(AC)) P(AUBUC)=.55+.5...?
anonymous
  • anonymous
im not sure what that last term would be though
amistre64
  • amistre64
yeah, im coming up empty on it as well
anonymous
  • anonymous
here is what i have so far p(A)=P(B)=B(C)=.5 p(AB)=.45 p(AC)=.2 P(BC)=.3
anonymous
  • anonymous
does this look right?
anonymous
  • anonymous
for P(AC) im getting .3
anonymous
  • anonymous
yeah my mistake
anonymous
  • anonymous
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
anonymous
  • anonymous
somehow we need two expressions that contain p(ABC) that we know everything except that one
anonymous
  • anonymous
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)
anonymous
  • anonymous
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?
anonymous
  • anonymous
yeah we get that for sure it is \(.45\)
anonymous
  • anonymous
i mean it is \(.45-x\)
anonymous
  • anonymous
putting \(x=P(ABC)\)
anonymous
  • anonymous
now we need some other expression involving \(x\)
anonymous
  • anonymous
see i was thinking if you treat AUB = y then could you find P(yUC)? which would be P(AUBUC)
anonymous
  • anonymous
i am thinking some demorgan law
anonymous
  • anonymous
maybe finding \[P(B^cC^c)\]?
anonymous
  • anonymous
am i being stupid, or does it say \(x=2(1-x)\)?
anonymous
  • anonymous
if \(P(ABC)=x\), then \(P((ABC)^c)=1-x\)
anonymous
  • anonymous
oh no
anonymous
  • anonymous
i guess i read it incorrectly
anonymous
  • anonymous
its \[P(ABC) = 2(P(ABC^c))\]
anonymous
  • anonymous
just the complement of C
anonymous
  • anonymous
oooh ok let me look at my diagram again
anonymous
  • anonymous
so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?
anonymous
  • anonymous
|dw:1358826338804:dw|
anonymous
  • anonymous
an we know 2x+x=.45
anonymous
  • anonymous
so x=.15. but it should be .30
anonymous
  • anonymous
if i am reading the question correctly (the second time) it is not so bad
anonymous
  • anonymous
why?
anonymous
  • anonymous
because the answer in the back of the book says it should be haha
anonymous
  • anonymous
ok let me look again but i don't see a mistake yet
anonymous
  • anonymous
i have to keep scrolling up it is a pain maybe i will copy here
anonymous
  • anonymous
ya no problem
anonymous
  • anonymous
P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3
anonymous
  • anonymous
.55 = .5 + .5 - .45 so P(AB)=.45
anonymous
  • anonymous
P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)
anonymous
  • anonymous
i get \(3P(ABC)=.45\implies P(ABC)=.15\)
anonymous
  • anonymous
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.
anonymous
  • anonymous
there is a mistake here somewhere
anonymous
  • anonymous
what is P(AC)?
anonymous
  • anonymous
im getting .3 for P(AC)
anonymous
  • anonymous
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
anonymous
  • anonymous
wait i think i see what is wrong
anonymous
  • anonymous
\[P(ABC)=2P(ABC^c)\] .45=P(AB) .45+P(ABC)+P(ABC^c)
anonymous
  • anonymous
that + should be an = on the bottom one
anonymous
  • anonymous
yeah i got that
anonymous
  • anonymous
so wouldent it be .45/1.5?
anonymous
  • anonymous
oh crap i did it backwards!!
anonymous
  • anonymous
because we will have one P(ABC) and one P(ABC^c) but the equation calls for 2 P(ABC^c)'s ?
anonymous
  • anonymous
lol is that sound logic or am i making stuff up?
anonymous
  • anonymous
|dw:1358827302867:dw|
anonymous
  • anonymous
damn sometimes i am dense
anonymous
  • anonymous
no i think it was the other way? ill draw it real quick
anonymous
  • anonymous
really tyro error there, i did it backwards you get \(x=.15, 2x=.3\)
anonymous
  • anonymous
it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!
anonymous
  • anonymous
that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)
anonymous
  • anonymous
AHHHHH lol
anonymous
  • anonymous
got it!
anonymous
  • anonymous
jesus sometimes i think i should lose my math license
anonymous
  • anonymous
damn problem was not that hard only needed P(AB)
anonymous
  • anonymous
haha ya lol wow ok this problem was very frustrating
anonymous
  • anonymous
|dw:1358827656432:dw|
anonymous
  • anonymous
it has tripped up everyone i know so far that has tried it and its was a basic multiplication
anonymous
  • anonymous
yeah pretty basic, now that it is finished
anonymous
  • anonymous
lol well hopefully i will be able to recognize this stuff in the future. thank you for your help
anonymous
  • anonymous
yw

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