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Thread: Math problems

  1. #1
    Saizou is offline Senior Member Always Around
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    Default Math problems

    I seem to remember that we had a fun thread like this a while ago. The rules are simple, someone posts a problem, someone else solves it and posts a new one. Fun and educational, no?

    I'll start: A group of 2n boys and 2n girls is divided into two parts. What is the probability that both parts consist of an equal number of boys and girls?

  2. #2
    shautieh's Avatar
    shautieh is offline Senior Member Community Builder
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    It seems I am not the only one to dislike probas ^^

    Anyway, a possible answer is 100% if the 2 parts were made well !
    Of course it's another problem if they were divided into 2 parts randomly
    Seriously, it's a shame but I am too bad with probas :P
    And I am upping this thread as it could get interesting indeed !


  3. #3
    sakura_hana is offline Senior Member Community Builder
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    ^ lol ^__^
    i say 20%.thatis...if the parts do not consist of equal numbers of ppl.

    solve this:
    you're in a competition and you've just overtaken the last competitor. In what position are you now?

  4. #4
    shautieh's Avatar
    shautieh is offline Senior Member Community Builder
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    ^^

    Where do you get your 20% from ? It may be, but I doubt the probability is constant so there should be an "n" somewhere in the solution...

    As for your question there is no solution as it could be anyone from the first down to the "next to last" if if the race is circular, or none at all if it isn't.

  5. #5
    sakura_hana is offline Senior Member Community Builder
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    ooh wait, i misread that problem lol. 2n. didn't get the n part . and i was like ? wtf O__o
    is divided into two parts
    the question is: are those 2 parts equal?
    as for th other question,the answer is correct.
    Last edited by sakura_hana; 01-28-2008 at 03:34 PM.

  6. #6
    KyubiNoKitsune is offline Senior Member Respected Member
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    Better wait for Saizou's validation before posting a new problem?
    My guess is (0.5^n)%.
    I would suggest to post detailed solutions too. If my guess is right, then I'm going to write the explanation I had in mind.

  7. #7
    shautieh's Avatar
    shautieh is offline Senior Member Community Builder
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    @sakura : the 2 parts are not necessary equals as it was not specified, so I assume the only requirement is that there must be at least one person in each part.

    I also think we should wait for Saizou's validation before posting new problems

  8. #8
    Saizou is offline Senior Member Always Around
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    Well, this is what I get for not being specific enough. The parts are equal, which is actually stated in the question. How else could you get as many boys and girls in both groups?

    Also the selection is random, which was also implied, but not stated outright.

    And btw, the answer to the problem isn't (0.5^n). Detailed solutions would be good, and I'm going to make a detailed explanation of the answer anyway.

    And I also propose that if no one gets the right answer in, say five days, the solution is posted and a new problem is given. What do you all think?

  9. #9
    shautieh's Avatar
    shautieh is offline Senior Member Community Builder
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    ^ I thought you meant there should be the same number of men and women in each group (I assumed "independantly"), so it could be (for n = 2) 1 man and 1 woman in the first group, and 3 men and 3 women in the second.

    I also think the solution should be given after a couple of days so we can get another problem and the topic won't die.

  10. #10
    adonai is offline Senior Member Community Builder
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    ^ They're not independent, that's the problem.

    Google has failed me in my search for an algorithm.

 

 
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