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INT

  • 3 years ago

There is a 0.63 probability that Mobil stock will rise, a 0.75 probability Gulf stock will rise and a 0.4 chance that both stocks will rise. Find the probability that neither stock will rise. There is a 0.63 probability that Mobil stock will rise, a 0.75 probability Gulf stock will rise and a 0.4 chance that both stocks will rise. Find the probability that neither stock will rise. @Mathematics

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  1. INT
    • 3 years ago
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    So .37*.25=.0925 But the answer key for a similar problem( only difference is .47 instead of .4), the answer is .2775. The answers shouldnt differ that much right?

  2. Zarkon
    • 3 years ago
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    why are you multiplying those two numbers?

  3. INT
    • 3 years ago
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    isnt that how you get the probability for AND?

  4. Zarkon
    • 3 years ago
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    not in this case

  5. INT
    • 3 years ago
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    Then Im completely lost.

  6. Zarkon
    • 3 years ago
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    your events are NOT independent

  7. INT
    • 3 years ago
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    I think that's the concept that I dont quite understand

  8. INT
    • 3 years ago
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    would the formula be P(A U B)/P(A)?

  9. Zarkon
    • 3 years ago
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    no

  10. Zarkon
    • 3 years ago
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    use \[P(A^c\cap B^c)=1-P(A\cup B)\]

  11. INT
    • 3 years ago
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    so the 1-P(none) formula

  12. Zarkon
    • 3 years ago
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    for any event E \[P(E)=1-P(E^c)\] or \[P(E^c)=1-P(E)\]

  13. INT
    • 3 years ago
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    so how exactly does not being independent effect the problem?

  14. Zarkon
    • 3 years ago
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    events A and B are independent iff \[P(A\cap B)=P(A)P(B)\]

  15. Zarkon
    • 3 years ago
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    you cant use this formula since A and B are not independent

  16. INT
    • 3 years ago
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    So in terms of a venn diagram, independent is two circles that dont intersect. and not independent has two circles that do intersect. and in this problem the intersection is .4? So I need to find the sum of each part of the venn diagram, subtract from one (to get the number outside the diagram) which would be the "neither". Correct?

  17. Zarkon
    • 3 years ago
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    NO

  18. Zarkon
    • 3 years ago
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    independent events intersect (most of the time)

  19. Zarkon
    • 3 years ago
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    if A and B are independent and P(A)>0 and P(B)>0 then A and B have to intersect.

  20. Zarkon
    • 3 years ago
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    use \[P(A^c\cap B^c)=1-P(A\cup B)=1-[P(A)+P(B)-P(A\cap B)]\]

  21. INT
    • 3 years ago
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    |dw:1321204843404:dw| Is what Im picturing Basically, I was doing P(Ac∩Bc)=1−P(A∪B)=1−[P(A)+P(B)−P(A∩B)] except that mine adds instead of subtracts the intersection.

  22. INT
    • 3 years ago
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    so to get neither dont you need to get rid of all that is in the circles? including the intersection?

  23. Zarkon
    • 3 years ago
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    yes

  24. INT
    • 3 years ago
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    wouldnt that be adding the intersection? so 1−[P(A)+P(B)+P(A∩B)] instead of 1−[P(A)+P(B)−P(A∩B)]

  25. INT
    • 3 years ago
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    oh wait

  26. INT
    • 3 years ago
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    I see nevermind

  27. INT
    • 3 years ago
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    |dw:1321205205257:dw| So the answer is .02? that's seems really small.(compared to the other answer)

  28. Zarkon
    • 3 years ago
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    it is .02

  29. INT
    • 3 years ago
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    ok thanks

  30. Zarkon
    • 3 years ago
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    I would check your key again...if you just change from .4 to .47 then .2775 is not the answer

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