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

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- 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

hmm, auc = a + c - ac aub = a + b - ab bc = b (c|b) hmm

- anonymous

ya i found P(ab) to be .45 and P(AC) to be .3

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## More answers

- 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

using the first two you said

- amistre64

ac = .3 ab = .45

- amistre64

bc = .5 (c|b) = .3 c|b = 3/5

- 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

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- 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

correct im following that

- amistre64

im thinking if we knew c|a: p(ac) = a (c|a)

- anonymous

oh remember P(ABC)=2P(ABC^complement))

- amistre64

right, that most likely allows us to determine the universal set as a whole

- amistre64

.3 = .5 (c|a) .6 = c|a 4+5 = 60% of a as well; .3

- amistre64

ab = a (b|a) .45 = .5 (b|a) .9 = b|a such that 2+5 = .9*.5 = .45

- amistre64

might just have to wolf it to be certain :)

- anonymous

wolfram alpha?

- amistre64

yeah, but i dont know the syntax to write it up

- anonymous

ah ya me neither i dident know it could do probabilitys

- amistre64

there are those on here which can do these in their sleep, im just not one of them :)

- anonymous

haha alright i understand i cant either

- 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

see what i though was maybe you can do something like: P(AUB)UP(C)?

- anonymous

so P(AUBUC)=P(AUB)+P(C)-(P(AB)UP(AC)) P(AUBUC)=.55+.5...?

- anonymous

im not sure what that last term would be though

- amistre64

yeah, im coming up empty on it as well

- 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

does this look right?

- anonymous

for P(AC) im getting .3

- anonymous

yeah my mistake

- 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

somehow we need two expressions that contain p(ABC) that we know everything except that one

- 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

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

yeah we get that for sure it is \(.45\)

- anonymous

i mean it is \(.45-x\)

- anonymous

putting \(x=P(ABC)\)

- anonymous

now we need some other expression involving \(x\)

- anonymous

see i was thinking if you treat AUB = y then could you find P(yUC)? which would be P(AUBUC)

- anonymous

i am thinking some demorgan law

- anonymous

maybe finding \[P(B^cC^c)\]?

- anonymous

am i being stupid, or does it say \(x=2(1-x)\)?

- anonymous

if \(P(ABC)=x\), then \(P((ABC)^c)=1-x\)

- anonymous

oh no

- anonymous

i guess i read it incorrectly

- anonymous

its \[P(ABC) = 2(P(ABC^c))\]

- anonymous

just the complement of C

- anonymous

oooh ok let me look at my diagram again

- anonymous

so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?

- anonymous

|dw:1358826338804:dw|

- anonymous

an we know 2x+x=.45

- anonymous

so x=.15. but it should be .30

- anonymous

if i am reading the question correctly (the second time) it is not so bad

- anonymous

why?

- anonymous

because the answer in the back of the book says it should be haha

- anonymous

ok let me look again but i don't see a mistake yet

- anonymous

i have to keep scrolling up it is a pain maybe i will copy here

- anonymous

ya no problem

- anonymous

P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3

- anonymous

.55 = .5 + .5 - .45 so P(AB)=.45

- anonymous

P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)

- anonymous

i get \(3P(ABC)=.45\implies P(ABC)=.15\)

- 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

there is a mistake here somewhere

- anonymous

what is P(AC)?

- anonymous

im getting .3 for P(AC)

- 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

wait i think i see what is wrong

- anonymous

\[P(ABC)=2P(ABC^c)\] .45=P(AB) .45+P(ABC)+P(ABC^c)

- anonymous

that + should be an = on the bottom one

- anonymous

yeah i got that

- anonymous

so wouldent it be .45/1.5?

- anonymous

oh crap i did it backwards!!

- anonymous

because we will have one P(ABC) and one P(ABC^c) but the equation calls for 2 P(ABC^c)'s ?

- anonymous

lol is that sound logic or am i making stuff up?

- anonymous

|dw:1358827302867:dw|

- anonymous

damn sometimes i am dense

- anonymous

no i think it was the other way? ill draw it real quick

- anonymous

really tyro error there, i did it backwards you get \(x=.15, 2x=.3\)

- anonymous

it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!

- anonymous

that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)

- anonymous

AHHHHH lol

- anonymous

got it!

- anonymous

jesus sometimes i think i should lose my math license

- anonymous

damn problem was not that hard only needed P(AB)

- anonymous

haha ya lol wow ok this problem was very frustrating

- anonymous

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- anonymous

it has tripped up everyone i know so far that has tried it and its was a basic multiplication

- anonymous

yeah pretty basic, now that it is finished

- anonymous

lol well hopefully i will be able to recognize this stuff in the future. thank you for your help

- anonymous

yw

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