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Math Help - Probability (Pets)

  1. #1
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    Probability (Pets)

    Just started S1, just wondering if you could help me out with this question?

    Many thanks

    500 people own a fish, dog or cat. Of this, 314 own a fish, 228 own a dog and 312 own a cat. of these, 160 own a fish and a dog, 100 own a dog and a cat, 134 own a cat and a fish. calculate the number which own all 3 pets
    Then

    By using set notation, express these 3 events as at least two of these events occurring.
    tried:

    500 = F + (C-(C n F)) + (D - (D n F)) - (D n C) + (ALL3)

    This gives me a silly answer so must be wrong
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  2. #2
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    500=F+C+D-FC-FD-CD+FCD that gives the number owning at least one of the three.
    Last edited by Plato; October 23rd 2010 at 01:09 PM.
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  3. #3
    MHF Contributor harish21's Avatar
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    Denote fishes by F, dogs by D, and cats by C.


    for no.1, express the information given to you in set notation. Venn diagram will be useful. You have to show your work on where you are getting stuck.

    Your second part can be expressed as

     (D \cap C) \cup (D \cap F) \cup (C \cap F)
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  4. #4
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    thanks for the help, managed to get 44 for the first part.

    how do you express exactly two of F, C, D occurs?

    im thinking (F intersection C) intersection ( (F intersection D) union ( C intersection D) )
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Hm... I got 40 for the first part...

    Of if you like, I made 7 equations with 7 variables. From my Venn diagram, I labeled the parts a to g where e is the intersection af all three groups.

    a+b+c+d+e+f+g = 500, a+b+d+e = 314, b+c+e+f = 228, d+e+f+g = 312, b+e = 160, e+f = 100, d+e = 134 - Wolfram|Alpha
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  6. #6
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    Thanks for the 40, managed to find the mistake!

    Quote Originally Posted by Unknown008 View Post
    Hm... I got 40 for the first part...

    Of if you like, I made 7 equations with 7 variables. From my Venn diagram, I labeled the parts a to g where e is the intersection af all three groups.

    a+b+c+d+e+f+g = 500, a+b+d+e = 314, b+c+e+f = 228, d+e+f+g = 312, b+e = 160, e+f = 100, d+e = 134 - Wolfram|Alpha
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by steve13 View Post

    how do you express exactly two of F, C, D occurs?

    im thinking (F intersection C) intersection ( (F intersection D) union ( C intersection D) )
    NO that is not correct. Exactly two events occur means that two events occur and the third does not:

    (F \cap C \cap D^{c}) \cup (..........) \cup (..........)

    Can you finish?
    Last edited by harish21; October 24th 2010 at 11:07 AM.
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