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Probability and combinatrics with words

  1. Apr 26, 2012 #1
    4. An experiment consists of randomly rearranging the 9 letters of the word
    TARANTULA into a sequence of 9 letters, where all possible orders of these 9 letters are equally
    likely. Find the probability of each of the following events:

    (a) the first three letters include no A's;
    (b) the first three letters or the last three letters (or both) include no A's;
    (c) the fourth letter is the first A;
    (d) the first letter and the last letter are the same;
    (e) the word `TARANTULA' is obtained;
    (f ) the sequence contains the word `RAT'.


    Attempt at solutions :

    a) 6/9 * 5/8 * 4/7 (Probability of first non A * another non A letter * another non A letter)

    b) 6/9 * 5/8 * 4/7 + ( I am stuck I don't get it - I think it might just be 2 * anser of part a but I'm unable to think it through, I know though that if first 3 and last 3 have no A's, then middle will have all A's so it becomes more weird )

    c) 6/9 * 5/8 * 4/7 * 3/6 ( Prob. in a * probability of 4th letter being A in the scenario of question a)

    d) 5/9 * 1/8 + 5/9 * 2/8 (Probability of choosing a repeatable letter * probability of second letter coming up for both T and A since R U L N are not repeatable)

    e) 1/9!

    f) After working it out on paper it seems there are 6 permutations of RAT so 6 * 1/9!



    I am still not sure if these are correct and I know there must be a better way of doing these using combination formula so I would be greatful if someone could help me out.

    Thanks!
     
    Last edited: Apr 26, 2012
  2. jcsd
  3. Apr 27, 2012 #2
    Anyone? Please this is from a past paper that I am attempting since I have an exam on monday.
     
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