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Thread: Probability of Word within Word

  1. #1
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    Probability of Word within Word




    Hello!

    I'm in dire need of help as I've been working on this for a couple days now. I've been to the help center around school and have had three separate Prof. and student tutors give my a completely wrong answer (I checked with the Prof.). We received a problem with two pages of hints and explanations for how to approach but I still can't wrap my head around it. Here's the problem...

    Each letter of the word MATHEMATICS is placed on a card and put in a hat. What is the probability that the you pull the word MATH out by choosing 4 cards without replacement?

    The question doesn't clarify if order is necessary so I'd like to know both possibilities (that AHMT, MHTA and other variants still spell MATH, and I'd also like to figure out what the probability of drawing an M on the first, A on the second, T on the third, and an H).

    I initially approached this problem by trying to figure out how many ways there are to spell MATH. M as first, and so on. M=2, A=2, T=2 so 2^3 ways or 8 ways to draw. This would be the numerator.

    To solve for the other way for possibilities of spelling MATH I figure 4x3x2x1 (for the four letters in any order) which equals 24, one of which is exactly MATH.

    The denominator is where I'm having a bit of trouble. I thought it would be a permutation of 11x10x9x8 but realized both M's, A's, and T's are treated the same in spelling. So given that there are 2 of three letters, I did a permutation of 8x7x6x5 plus the number of ways double letters could be used which ended up being (MM)AAx4, (TT)MMx4, (AA)TTx4, plus double letters without double letters (if that makes sense) ((MMx7x6)x4), ((AAx7x6)x4), ((TTx7x6)x4). I multiplied by four to indicate that the position of the double letter could be varied (MM**, *MM*, **MM, M**M). So all together the numbers add up to be 1680 for the permutation of MATHEICS, 12 for the double double combinations, and 504 for the double single combinations to equal 2196.

    So my initial answer is 8/2196 or a probability of .003642.

    For the other way it would be 24/2196 for a probability of .010929.

    Can anyone help me out?

  2. #2
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    I get 8/7920 as the probability of drawing the word math in the order M - A - T - H, and without replacing the letters back into the bag each time. But that's what you thought at first so I must be missing something, and I'll admit I don't understand what you mean by double letters... If you draw two M's, that doesn't spell math, so why consider that?

  3. #3
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    11x10x9x8=7920

    That would be true if all letters were different. But IHM(1)E is the same as IHM(2)E. It's the same combination type even though different M's are used. So instead of 2 ways there is only 1. At least, that's what I got from the professor.

    By double letters I mean taking into account that there are 2 M's, 2 T's, and 2 A's in all the possibilities.
    Last edited by Abbott54; 04-16-2008 at 11:35 PM.

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