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Math Help - Permutations and Combinations in Probability

  1. #1
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    Permutations and Combinations in Probability

    Hi, can someone please explain what to do, stats really isn't my thing, so if you could please explain your answer, I'd really appreciate it!

    3 red cubes, 4 blue, 6 yellow are arranged in a row. find the probability that (a) the cubes at each end are the same color
    and (b) the cubes at each end are of a different color.

    Thanks a lot!
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  2. #2
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    Quote Originally Posted by overduex View Post
    Hi, can someone please explain what to do, stats really isn't my thing, so if you could please explain your answer, I'd really appreciate it!
    3 red cubes, 4 blue, 6 yellow are arranged in a row. find the probability that (a) the cubes at each end are the same color
    The total number of ways the arrange thes cubes is N=\frac{13!}{(3!)(4!)(6!)} that becomes your denominator.
    The number of ways to have a reds at each end is \frac{11!}{(1!)(4!)(6!)}

    Now you do it for the other two colors.
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  3. #3
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    Quote Originally Posted by Plato View Post
    The total number of ways the arrange thes cubes is N=\frac{13!}{(3!)(4!)(6!)} that becomes your denominator.
    The number of ways to have a reds at each end is \frac{11!}{(1!)(4!)(6!)}

    Now you do it for the other two colors.
    after you do it for the other 2 colors, do you multipy them together? or add?
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  4. #4
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    Quote Originally Posted by overduex View Post
    after you do it for the other 2 colors, do you multipy them together? or add?
    ADD
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  5. #5
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    Quote Originally Posted by Plato View Post
    ADD
    sorry to bother again, but could you explain how to get the answer?
    It is supposed to be 4/13 but i keep getting ridiculous numbers as answers. I basically just added what you had written
    11!/(1!)(4!)(6!) + 11!/(3!)(2!)(6!) + 11!/(3!)(4!)(4!)

    do you see anything I may be doing wrong?
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  6. #6
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    Iunno whats wrong. I tried it out, I i seemed to get 4/13. Maybe you're simplifying wrong?

    11!/(1!x4!x6!) = 2310

    11!/(3!x2!x6!) = 4620

    11!/(3!x4!x4!) = 11550

    13!/(3!x4!x6!) = 60060

    So, (2310/60060) + (4620/60060) + (11550/60060) = (18480/60060) = (4/13).

    If your entering the numbers into your calculator like this 11!/3!x4!x4!, then you might be getting the wrong answer, be in reality it'd be this:
    (11!/3!) x 4! x 4! which might explain the random numbers.
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  7. #7
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    Quote Originally Posted by zerobladex View Post
    Iunno whats wrong. I tried it out, I i seemed to get 4/13. Maybe you're simplifying wrong?

    11!/(1!x4!x6!) = 2310

    11!/(3!x2!x6!) = 4620

    11!/(3!x4!x4!) = 11550

    13!/(3!x4!x6!) = 60060

    So, (2310/60060) + (4620/60060) + (11550/60060) = (18480/60060) = (4/13).

    If your entering the numbers into your calculator like this 11!/3!x4!x4!, then you might be getting the wrong answer, be in reality it'd be this:
    (11!/3!) x 4! x 4! which might explain the random numbers.

    yeah, I figured out my mistake. I was wayyy off. haha, thanks for trying though!
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