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Simple coin tossing question, confused on their answer

  1. Nov 10, 2006 #1
    Hello everyone. I"m just wondering why they solved the answer in this manner.

    The question is:
    A coin is tossed 4 times. Each time the result H for heads or T for tails is recorded. An outcome of HHTT means that heads were obtained on the first 2 tosses and tails on the second 2. Assume heads and tails are equally like on each toss.

    Well they solved it this way:
    There are 4 outcomes in which exactly one head can occur (since a string of one "T" and three "H"'s can have
    the "T" in any one of the string's four positions). So the probability of exactly one head is 4/2^4 = 1/4.

    Okay I understand that there are 2^4 possible outcomes on 4 tosses, 2 chocies can happen, either a H or a T, so 2^4.

    But Why did they say, since a string of one "T" and 3 "H"'s. If we are trying to find how many times you get exactly 1 head Why wouldn't they say the following:
    since a string of one "H" and three "T"s can have the "H" in any one of the strings for positions, such as HTTT, THTT, TTHT, TTTH, this shows its 4 ways, in which you will get exactly 1 head. I'm just confused on why they worded it that way or am I missing somthing?

    Thanks
     
  2. jcsd
  3. Nov 10, 2006 #2

    quasar987

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    It's probably just an error in the book.
     
  4. Nov 10, 2006 #3
    I thought so, but just wanted to make sure, the professor actually did the solution but either way it would come out to the same answer just making sure. thanks!
     
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