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Another Probability Question(Not so simple)

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data

    An honest coin is throughn again and again until two consigutive heads or tails appears.
    Assume that the troughs are independent. Let T be the number of throughs which can take on the values 2,3,....

    Calculate [tex]P(T = n)[/tex] for [tex]n = 2,3,...[/tex]


    3. The attempt at a solution

    P(HH,TT) = P(HH) + P(TT) = [tex] 2^{-n} + 2^{-n} = \frac{1}{2^{(-n-1)}}[/tex]

    for n = 2,3,....

    Does this look okay? Or do I need to add something more text?? If yes, could somebody please give me a hint or surgestion?

    Sincerely Yours
    Beowulf
     
    Last edited: Nov 26, 2007
  2. jcsd
  3. Nov 27, 2007 #2

    Shooting Star

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    Homework Helper

    (you have summed the final answer wrongly. Also, please use a spell-checker.)

    If I understand correctly, (n-1) throws have gone before the nth throw, with alternating H and T. The (n-1)th and the nth throw are HH or TT.

    If it had started with H, then it has gone like HTHT... for n-1 throws. The P of that is (1/2)^(n-1). The P of the nth throw matching that of n-1 is 1/2. So, total P for this case is (1/2)^(n-1)*(1/2).

    Similarly, if it had started with T, then the total P of nth being same as n-1 is again (1/2)^(n-1)*(1/2).

    So, the reqd P is the sum of the two = (1/2)^(n-1).

    This tallies with your answer.

    (As a check, let n=2, which means that the first two throws will be the same. We know that P of HH or TT is 1/2. Similarly, test for n=3. HTT or THH has prob 1/4.)
     
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