At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

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

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

using the first two you said

ac = .3
ab = .45

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

|dw:1358793002972:dw|

correct im following that

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

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

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

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

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

might just have to wolf it to be certain :)

wolfram alpha?

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

ah ya me neither i dident know it could do probabilitys

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

haha alright i understand i cant either

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

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

im not sure what that last term would be though

yeah, im coming up empty on it as well

here is what i have so far
p(A)=P(B)=B(C)=.5
p(AB)=.45
p(AC)=.2
P(BC)=.3

does this look right?

for P(AC) im getting .3

yeah my mistake

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

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

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

putting \(x=P(ABC)\)

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

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

i am thinking some demorgan law

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

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

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

oh no

i guess i read it incorrectly

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

just the complement of C

oooh ok let me look at my diagram again

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

|dw:1358826338804:dw|

an we know 2x+x=.45

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

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

why?

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

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

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

ya no problem

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

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

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

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

there is a mistake here somewhere

what is P(AC)?

im getting .3 for P(AC)

wait i think i see what is wrong

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

that + should be an = on the bottom one

yeah i got that

so wouldent it be .45/1.5?

oh crap i did it backwards!!

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

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

|dw:1358827302867:dw|

damn sometimes i am dense

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

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

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

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

AHHHHH lol

got it!

jesus sometimes i think i should lose my math license

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

haha ya lol wow ok this problem was very frustrating

|dw:1358827656432:dw|

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

yeah pretty basic, now that it is finished

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

yw