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Math Help - is hypergeomdist a correct choice?

  1. #1
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    is hypergeomdist a correct choice?

    I want to find the odds of something:

    I have 60 objects, of which I am going to select seven. Of these objects, 15 are the same (call them broken). What are the odds of me having at least three broken objects in my selection of seven?

    I intended to go about this along the lines of

    1-hypergeomdist(0, 7, 15, 60) -hypergeomdist(1, 7, 15, 60)-hypergeomdist(2, 7, 15, 60)

    Second question:

    Say 15 are broken in one way (A) and 7 are broken in another way (B). I now select ten of them. What are the odds of having at least two broken in one way and three broken in the other?

    Do I multiply these odds (assuming I'm setting the problems up correctly in the first place) just like I would simpler probabilities? At this point i'm assuming that none of the objects are broken in both ways at the same time.

    Thanks. Also, I've looked for sites dealing with the second question online and found little to nothing. Any sites you can direct me to so I can find my own answers. Not that I don't like giving you something to do. . .
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  2. #2
    MHF Contributor matheagle's Avatar
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    The first one is correct AND this a hypergeo as long as we are selecting without relacement.

    As for the second question, think of it as 3 groups.
    (Just brains, no sites.)
    You are selecting at least 2 from A, 3 from B, rest from C.
    To do that exactly, well thats...

    { {15\choose 2} {7\choose 3} {38\choose 5}\over {60\choose 10}}

    Next figure out all the possibilities, 3 from A, 3 from B, 4 from C....

    { {15\choose 3} {7\choose 3} {38\choose 4}\over {60\choose 10}}.

    FIND all of these and add them together. The complement here is a pain.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    The first one is correct AND this a hypergeo as long as we are selecting without relacement.

    As for the second question, think of it as 3 groups.
    (Just brains, no sites.)
    You are selecting at least 2 from A, 3 from B, rest from C.
    To do that exactly, well thats...

    { {15\choose 2} {7\choose 3} {38\choose 5}\over {60\choose 10}}

    Next figure out all the possibilities, 3 from A, 3 from B, 4 from C....

    { {15\choose 3} {7\choose 3} {38\choose 4}\over {60\choose 10}}.

    FIND all of these and add them together. The complement here is a pain.
    Can we subtract all other choices instead?

    1-({{15\choose 0}{7\choose0}{38\choose10}\over{60\choose10}}+{{15  \choose 1}{7\choose0}{38\choose9}\over{60\choose10}}+{{15\  choose 1}{7\choose1}{38\choose8}\over{60\choose10}}...)

    and is there anyway to do it multiplying our options of hyper geometric distributions, or am i just hoping here? cuz if we can do it that way, it would seem easier . . .
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  4. #4
    MHF Contributor matheagle's Avatar
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    I wouldn't look at the complement. If you did you'd have a mess.
    You can have 0 from group A, then 0 thru 7 from group B
    You can have 1 from group B, then 0 thru 7 from group B....
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  5. #5
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    Quote Originally Posted by matheagle View Post
    I wouldn't look at the complement. If you did you'd have a mess.
    You can have 0 from group A, then 0 thru 7 from group B
    You can have 1 from group B, then 0 thru 7 from group B....
    Both seems almost as bad as the other, but both are decent to set up in excel, i think. I'll mess around with it. My goal here would be to create something that would allow me to fluctuate both the total number of A and B broken in the set as well as a set number of acceptable of from both in the given selection.

    Food for thought, certainly, and thank you.
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