Benbburn
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))?
Delete
Share
This Question is Closed
Benbburn
Best Response
You'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
amistre64
Best Response
You've already chosen the best response.
0
hmm, auc = a + c - ac
aub = a + b - ab
bc = b (c|b)
hmm
Benbburn
Best Response
You've already chosen the best response.
0
ya i found P(ab) to be .45 and P(AC) to be .3
amistre64
Best Response
You'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
Benbburn
Best Response
You've already chosen the best response.
0
using the first two you said
amistre64
Best Response
You've already chosen the best response.
0
ac = .3
ab = .45
amistre64
Best Response
You've already chosen the best response.
0
bc = .5 (c|b) = .3
c|b = 3/5
Benbburn
Best Response
You'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)
amistre64
Best Response
You've already chosen the best response.
0
|dw:1358793002972:dw|
amistre64
Best Response
You'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
Benbburn
Best Response
You've already chosen the best response.
0
correct im following that
amistre64
Best Response
You've already chosen the best response.
0
im thinking if we knew c|a: p(ac) = a (c|a)
Benbburn
Best Response
You've already chosen the best response.
0
oh remember P(ABC)=2P(ABC^complement))
amistre64
Best Response
You've already chosen the best response.
0
right, that most likely allows us to determine the universal set as a whole
amistre64
Best Response
You've already chosen the best response.
0
.3 = .5 (c|a)
.6 = c|a
4+5 = 60% of a as well; .3
amistre64
Best Response
You've already chosen the best response.
0
ab = a (b|a)
.45 = .5 (b|a)
.9 = b|a such that 2+5 = .9*.5 = .45
amistre64
Best Response
You've already chosen the best response.
0
might just have to wolf it to be certain :)
Benbburn
Best Response
You've already chosen the best response.
0
wolfram alpha?
amistre64
Best Response
You've already chosen the best response.
0
yeah, but i dont know the syntax to write it up
Benbburn
Best Response
You've already chosen the best response.
0
ah ya me neither i dident know it could do probabilitys
amistre64
Best Response
You'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 :)
Benbburn
Best Response
You've already chosen the best response.
0
haha alright i understand i cant either
amistre64
Best Response
You'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)
Benbburn
Best Response
You've already chosen the best response.
0
see what i though was maybe you can do something like:
P(AUB)UP(C)?
Benbburn
Best Response
You've already chosen the best response.
0
so P(AUBUC)=P(AUB)+P(C)-(P(AB)UP(AC))
P(AUBUC)=.55+.5...?
Benbburn
Best Response
You've already chosen the best response.
0
im not sure what that last term would be though
amistre64
Best Response
You've already chosen the best response.
0
yeah, im coming up empty on it as well
anonymous
Best Response
You've already chosen the best response.
0
here is what i have so far
p(A)=P(B)=B(C)=.5
p(AB)=.45
p(AC)=.2
P(BC)=.3
anonymous
Best Response
You've already chosen the best response.
0
does this look right?
Benbburn
Best Response
You've already chosen the best response.
0
for P(AC) im getting .3
anonymous
Best Response
You've already chosen the best response.
0
yeah my mistake
Benbburn
Best Response
You'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
anonymous
Best Response
You've already chosen the best response.
0
somehow we need two expressions that contain p(ABC) that we know everything except that one
Benbburn
Best Response
You'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)
Benbburn
Best Response
You'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?
anonymous
Best Response
You've already chosen the best response.
0
yeah we get that for sure
it is \(.45\)
anonymous
Best Response
You've already chosen the best response.
0
i mean it is \(.45-x\)
anonymous
Best Response
You've already chosen the best response.
0
putting \(x=P(ABC)\)
anonymous
Best Response
You've already chosen the best response.
0
now we need some other expression involving \(x\)
Benbburn
Best Response
You'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)
anonymous
Best Response
You've already chosen the best response.
0
i am thinking some demorgan law
Benbburn
Best Response
You've already chosen the best response.
0
maybe finding \[P(B^cC^c)\]?
anonymous
Best Response
You've already chosen the best response.
0
am i being stupid, or does it say \(x=2(1-x)\)?
anonymous
Best Response
You've already chosen the best response.
0
if \(P(ABC)=x\), then \(P((ABC)^c)=1-x\)
Benbburn
Best Response
You've already chosen the best response.
0
oh no
anonymous
Best Response
You've already chosen the best response.
0
i guess i read it incorrectly
Benbburn
Best Response
You've already chosen the best response.
0
its \[P(ABC) = 2(P(ABC^c))\]
Benbburn
Best Response
You've already chosen the best response.
0
just the complement of C
anonymous
Best Response
You've already chosen the best response.
0
oooh ok let me look at my diagram again
anonymous
Best Response
You've already chosen the best response.
0
so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?
anonymous
Best Response
You've already chosen the best response.
0
|dw:1358826338804:dw|
Benbburn
Best Response
You've already chosen the best response.
0
an we know 2x+x=.45
Benbburn
Best Response
You've already chosen the best response.
0
so x=.15. but it should be .30
anonymous
Best Response
You've already chosen the best response.
0
if i am reading the question correctly (the second time) it is not so bad
anonymous
Best Response
You've already chosen the best response.
0
why?
Benbburn
Best Response
You've already chosen the best response.
0
because the answer in the back of the book says it should be haha
anonymous
Best Response
You've already chosen the best response.
0
ok let me look again but i don't see a mistake yet
anonymous
Best Response
You've already chosen the best response.
0
i have to keep scrolling up it is a pain
maybe i will copy here
Benbburn
Best Response
You've already chosen the best response.
0
ya no problem
anonymous
Best Response
You've already chosen the best response.
0
P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3
anonymous
Best Response
You've already chosen the best response.
0
.55 = .5 + .5 - .45 so P(AB)=.45
anonymous
Best Response
You've already chosen the best response.
0
P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)
anonymous
Best Response
You've already chosen the best response.
0
i get \(3P(ABC)=.45\implies P(ABC)=.15\)
Benbburn
Best Response
You'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.
anonymous
Best Response
You've already chosen the best response.
0
there is a mistake here somewhere
anonymous
Best Response
You've already chosen the best response.
0
what is P(AC)?
Benbburn
Best Response
You've already chosen the best response.
0
im getting .3 for P(AC)
anonymous
Best Response
You've already chosen the best response.
0
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
Benbburn
Best Response
You've already chosen the best response.
0
wait i think i see what is wrong
Benbburn
Best Response
You've already chosen the best response.
0
\[P(ABC)=2P(ABC^c)\]
.45=P(AB)
.45+P(ABC)+P(ABC^c)
Benbburn
Best Response
You've already chosen the best response.
0
that + should be an = on the bottom one
anonymous
Best Response
You've already chosen the best response.
0
yeah i got that
Benbburn
Best Response
You've already chosen the best response.
0
so wouldent it be .45/1.5?
anonymous
Best Response
You've already chosen the best response.
0
oh crap i did it backwards!!
Benbburn
Best Response
You'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
?
Benbburn
Best Response
You've already chosen the best response.
0
lol is that sound logic or am i making stuff up?
anonymous
Best Response
You've already chosen the best response.
0
|dw:1358827302867:dw|
anonymous
Best Response
You've already chosen the best response.
0
damn sometimes i am dense
Benbburn
Best Response
You've already chosen the best response.
0
no i think it was the other way? ill draw it real quick
anonymous
Best Response
You've already chosen the best response.
0
really tyro error there, i did it backwards
you get \(x=.15, 2x=.3\)
anonymous
Best Response
You've already chosen the best response.
0
it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!
anonymous
Best Response
You've already chosen the best response.
0
that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)
Benbburn
Best Response
You've already chosen the best response.
0
AHHHHH lol
Benbburn
Best Response
You've already chosen the best response.
0
got it!
anonymous
Best Response
You've already chosen the best response.
0
jesus sometimes i think i should lose my math license
anonymous
Best Response
You've already chosen the best response.
0
damn problem was not that hard
only needed P(AB)
Benbburn
Best Response
You've already chosen the best response.
0
haha ya lol wow ok this problem was very frustrating
anonymous
Best Response
You've already chosen the best response.
0
|dw:1358827656432:dw|
Benbburn
Best Response
You'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
anonymous
Best Response
You've already chosen the best response.
0
yeah pretty basic, now that it is finished
Benbburn
Best Response
You'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
anonymous
Best Response
You've already chosen the best response.
0
yw