## Benbburn 2 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))?

1. Benbburn

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

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

3. Benbburn

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

4. 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

5. Benbburn

using the first two you said

6. amistre64

ac = .3 ab = .45

7. amistre64

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

8. Benbburn

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

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10. 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

11. Benbburn

correct im following that

12. amistre64

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

13. Benbburn

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

14. amistre64

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

15. amistre64

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

16. amistre64

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

17. amistre64

might just have to wolf it to be certain :)

18. Benbburn

wolfram alpha?

19. amistre64

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

20. Benbburn

ah ya me neither i dident know it could do probabilitys

21. amistre64

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

22. Benbburn

haha alright i understand i cant either

23. 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)

24. Benbburn

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

25. Benbburn

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

26. Benbburn

im not sure what that last term would be though

27. amistre64

yeah, im coming up empty on it as well

28. satellite73

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

29. satellite73

does this look right?

30. Benbburn

for P(AC) im getting .3

31. satellite73

yeah my mistake

32. Benbburn

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. satellite73

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

34. Benbburn

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

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. satellite73

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

37. satellite73

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

38. satellite73

putting \(x=P(ABC)\)

39. satellite73

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

40. Benbburn

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

41. satellite73

i am thinking some demorgan law

42. Benbburn

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

43. satellite73

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

44. satellite73

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

45. Benbburn

oh no

46. satellite73

i guess i read it incorrectly

47. Benbburn

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

48. Benbburn

just the complement of C

49. satellite73

oooh ok let me look at my diagram again

50. satellite73

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

51. satellite73

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52. Benbburn

an we know 2x+x=.45

53. Benbburn

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

54. satellite73

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

55. satellite73

why?

56. Benbburn

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

57. satellite73

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

58. satellite73

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

59. Benbburn

ya no problem

60. satellite73

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

61. satellite73

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

62. satellite73

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

63. satellite73

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

64. Benbburn

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. satellite73

there is a mistake here somewhere

66. satellite73

what is P(AC)?

67. Benbburn

im getting .3 for P(AC)

68. satellite73

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

wait i think i see what is wrong

70. Benbburn

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

71. Benbburn

that + should be an = on the bottom one

72. satellite73

yeah i got that

73. Benbburn

so wouldent it be .45/1.5?

74. satellite73

oh crap i did it backwards!!

75. Benbburn

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

76. Benbburn

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

77. satellite73

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78. satellite73

damn sometimes i am dense

79. Benbburn

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

80. satellite73

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

81. satellite73

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

82. satellite73

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

83. Benbburn

AHHHHH lol

84. Benbburn

got it!

85. satellite73

jesus sometimes i think i should lose my math license

86. satellite73

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

87. Benbburn

haha ya lol wow ok this problem was very frustrating

88. satellite73

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89. Benbburn

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

90. satellite73

yeah pretty basic, now that it is finished

91. Benbburn

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

92. satellite73

yw