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Urgent: probability question

  1. Aug 24, 2005 #1
    hi, i posted this question in the probability forum here but noone is even browsing that one.

    my brother needs to know this but i'm in exam block right now and can only think of the physics im studying this term. i cant remember how to figure this out anymore i just know i'll probably have to use nCr and nPr.
    heres the question:
    how do i find the probability of getting tossing 3 heads in a row out of 64 tosses?

    thanks, much appreciated.
     
  2. jcsd
  3. Aug 24, 2005 #2

    SGT

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    Do you mean getting 3 heads and 61 tails or 3 heads in a row?
     
  4. Aug 25, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    My interpretation was getting 3 heads in a row (at least once or exactly once? Exactly three or would 4 heads in a row also count?) anywhere in the 61 flips. Looks to me like a very hard problem with a lot of calculation involved.
     
  5. Aug 25, 2005 #4

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    I will assume the problem is having at least 3 consecutive heads at least once.
    In 3 flips you have the following possibilities:
    TTT
    TTH
    THT
    THH
    HTT
    HTH
    HHT
    HHH
    So you have a 1 in 8 probability to get threee heads in a row with 3 flips.
    You have 62 groups of three consecutive flips in 64 tosses:
    1 2 3 - 2 3 4 - ... - 62 63 64.
    You have 1/8 probability of getting 3 heads in the first group and 7/8 of not getting it.
    The probability of getting the 3 heads in the second group given that you did not get it in the first is
    7/8 * 1/8 = 7/64
    So you have a 57/64 chance of not getting 3 heads in the first 2 groups.
    The probability of getting the 3 heads in the third group given that you did not get it in the first two is:
    57/64 * 1/8 = 57/512
    and so on...
    The total probability is the sum 1/8 + 7/64 + 57/512 + ... very near 1.
     
    Last edited by a moderator: Aug 25, 2005
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