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Benbburn

  • 3 years 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))?

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  1. Benbburn
    • 3 years ago
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    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

  2. amistre64
    • 3 years ago
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    hmm, auc = a + c - ac aub = a + b - ab bc = b (c|b) hmm

  3. Benbburn
    • 3 years ago
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    ya i found P(ab) to be .45 and P(AC) to be .3

  4. amistre64
    • 3 years ago
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    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

  5. Benbburn
    • 3 years ago
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    using the first two you said

  6. amistre64
    • 3 years ago
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    ac = .3 ab = .45

  7. amistre64
    • 3 years ago
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    bc = .5 (c|b) = .3 c|b = 3/5

  8. Benbburn
    • 3 years ago
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    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)

  9. amistre64
    • 3 years ago
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    |dw:1358793002972:dw|

  10. amistre64
    • 3 years ago
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    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

  11. Benbburn
    • 3 years ago
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    correct im following that

  12. amistre64
    • 3 years ago
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    im thinking if we knew c|a: p(ac) = a (c|a)

  13. Benbburn
    • 3 years ago
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    oh remember P(ABC)=2P(ABC^complement))

  14. amistre64
    • 3 years ago
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    right, that most likely allows us to determine the universal set as a whole

  15. amistre64
    • 3 years ago
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    .3 = .5 (c|a) .6 = c|a 4+5 = 60% of a as well; .3

  16. amistre64
    • 3 years ago
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    ab = a (b|a) .45 = .5 (b|a) .9 = b|a such that 2+5 = .9*.5 = .45

  17. amistre64
    • 3 years ago
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    might just have to wolf it to be certain :)

  18. Benbburn
    • 3 years ago
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    wolfram alpha?

  19. amistre64
    • 3 years ago
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    yeah, but i dont know the syntax to write it up

  20. Benbburn
    • 3 years ago
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    ah ya me neither i dident know it could do probabilitys

  21. amistre64
    • 3 years ago
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    there are those on here which can do these in their sleep, im just not one of them :)

  22. Benbburn
    • 3 years ago
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    haha alright i understand i cant either

  23. amistre64
    • 3 years ago
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    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)

  24. Benbburn
    • 3 years ago
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    see what i though was maybe you can do something like: P(AUB)UP(C)?

  25. Benbburn
    • 3 years ago
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    so P(AUBUC)=P(AUB)+P(C)-(P(AB)UP(AC)) P(AUBUC)=.55+.5...?

  26. Benbburn
    • 3 years ago
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    im not sure what that last term would be though

  27. amistre64
    • 3 years ago
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    yeah, im coming up empty on it as well

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

  29. anonymous
    • 3 years ago
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    does this look right?

  30. Benbburn
    • 3 years ago
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    for P(AC) im getting .3

  31. anonymous
    • 3 years ago
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    yeah my mistake

  32. Benbburn
    • 3 years ago
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    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

  33. anonymous
    • 3 years ago
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    somehow we need two expressions that contain p(ABC) that we know everything except that one

  34. Benbburn
    • 3 years ago
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    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)

  35. Benbburn
    • 3 years ago
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    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?

  36. anonymous
    • 3 years ago
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    yeah we get that for sure it is \(.45\)

  37. anonymous
    • 3 years ago
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    i mean it is \(.45-x\)

  38. anonymous
    • 3 years ago
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    putting \(x=P(ABC)\)

  39. anonymous
    • 3 years ago
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    now we need some other expression involving \(x\)

  40. Benbburn
    • 3 years ago
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    see i was thinking if you treat AUB = y then could you find P(yUC)? which would be P(AUBUC)

  41. anonymous
    • 3 years ago
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    i am thinking some demorgan law

  42. Benbburn
    • 3 years ago
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    maybe finding \[P(B^cC^c)\]?

  43. anonymous
    • 3 years ago
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    am i being stupid, or does it say \(x=2(1-x)\)?

  44. anonymous
    • 3 years ago
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    if \(P(ABC)=x\), then \(P((ABC)^c)=1-x\)

  45. Benbburn
    • 3 years ago
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    oh no

  46. anonymous
    • 3 years ago
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    i guess i read it incorrectly

  47. Benbburn
    • 3 years ago
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    its \[P(ABC) = 2(P(ABC^c))\]

  48. Benbburn
    • 3 years ago
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    just the complement of C

  49. anonymous
    • 3 years ago
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    oooh ok let me look at my diagram again

  50. anonymous
    • 3 years ago
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    so \(P(ABC^c)=2x\) so why isn't t \(3x=.45\) ?

  51. anonymous
    • 3 years ago
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    |dw:1358826338804:dw|

  52. Benbburn
    • 3 years ago
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    an we know 2x+x=.45

  53. Benbburn
    • 3 years ago
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    so x=.15. but it should be .30

  54. anonymous
    • 3 years ago
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    if i am reading the question correctly (the second time) it is not so bad

  55. anonymous
    • 3 years ago
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    why?

  56. Benbburn
    • 3 years ago
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    because the answer in the back of the book says it should be haha

  57. anonymous
    • 3 years ago
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    ok let me look again but i don't see a mistake yet

  58. anonymous
    • 3 years ago
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    i have to keep scrolling up it is a pain maybe i will copy here

  59. Benbburn
    • 3 years ago
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    ya no problem

  60. anonymous
    • 3 years ago
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    P(ABC) if P(a)=P(B)=P( C)=.5, P(AUB)=.55, P(AUC)=.7, P(BC)=.3

  61. anonymous
    • 3 years ago
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    .55 = .5 + .5 - .45 so P(AB)=.45

  62. anonymous
    • 3 years ago
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    P(AB) = P(ABC) + P(ABC^c) as the sets are disjoint and there union is \(AB\)

  63. anonymous
    • 3 years ago
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    i get \(3P(ABC)=.45\implies P(ABC)=.15\)

  64. Benbburn
    • 3 years ago
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    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.

  65. anonymous
    • 3 years ago
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    there is a mistake here somewhere

  66. anonymous
    • 3 years ago
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    what is P(AC)?

  67. Benbburn
    • 3 years ago
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    im getting .3 for P(AC)

  68. anonymous
    • 3 years ago
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    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

  69. Benbburn
    • 3 years ago
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    wait i think i see what is wrong

  70. Benbburn
    • 3 years ago
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    \[P(ABC)=2P(ABC^c)\] .45=P(AB) .45+P(ABC)+P(ABC^c)

  71. Benbburn
    • 3 years ago
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    that + should be an = on the bottom one

  72. anonymous
    • 3 years ago
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    yeah i got that

  73. Benbburn
    • 3 years ago
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    so wouldent it be .45/1.5?

  74. anonymous
    • 3 years ago
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    oh crap i did it backwards!!

  75. Benbburn
    • 3 years ago
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    because we will have one P(ABC) and one P(ABC^c) but the equation calls for 2 P(ABC^c)'s ?

  76. Benbburn
    • 3 years ago
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    lol is that sound logic or am i making stuff up?

  77. anonymous
    • 3 years ago
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    |dw:1358827302867:dw|

  78. anonymous
    • 3 years ago
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    damn sometimes i am dense

  79. Benbburn
    • 3 years ago
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    no i think it was the other way? ill draw it real quick

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

  81. anonymous
    • 3 years ago
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    it is \(P(ABC)=2P(ABC^c)\) i.e. P(ABC\) is twice as big!!

  82. anonymous
    • 3 years ago
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    that is why \(P(ABC)=.3\) as in the book and \(P(ABC^c)=.15\)

  83. Benbburn
    • 3 years ago
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    AHHHHH lol

  84. Benbburn
    • 3 years ago
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    got it!

  85. anonymous
    • 3 years ago
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    jesus sometimes i think i should lose my math license

  86. anonymous
    • 3 years ago
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    damn problem was not that hard only needed P(AB)

  87. Benbburn
    • 3 years ago
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    haha ya lol wow ok this problem was very frustrating

  88. anonymous
    • 3 years ago
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    |dw:1358827656432:dw|

  89. Benbburn
    • 3 years ago
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    it has tripped up everyone i know so far that has tried it and its was a basic multiplication

  90. anonymous
    • 3 years ago
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    yeah pretty basic, now that it is finished

  91. Benbburn
    • 3 years ago
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    lol well hopefully i will be able to recognize this stuff in the future. thank you for your help

  92. anonymous
    • 3 years ago
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    yw

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