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Probabilty of one tail followed by three heads with a biased coin

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A coin is biased so that the probability of heads is 2/3. What is the probability that exactly four heads come up when the coin is flipped seven times, assuming that the flips are independent?


    2. Relevant equations



    3. The attempt at a solution
    C(7,4)(2/3)^4(1/3)^3 = (35*16)/3^7 = 560/2187

    Now, I know this answer is correct, but what if we were asked to find probability of exactly one tail followed by three heads rather than exactly four heads: How would I calculate the answer for that?

    Thank you.
     
  2. jcsd
  3. Feb 11, 2013 #2

    Ray Vickson

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    How would you calculate the probability of getting T on toss 1? If your first toss is T, does that affect the probability that the next three tosses all give H?
     
  4. Feb 11, 2013 #3
    No, it doesn't since each toss is independent. So, it would be calculated the same way?
     
  5. Feb 12, 2013 #4

    haruspex

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    The four heads were out of 7 tosses. Are your 1T3H out of just four tosses? If not, what do you mean?
     
  6. Feb 12, 2013 #5
    I believe he was referencing another problem he posted on PF. But yeah, I assume that THHH would be out of four tosses; obviously any later ones would be superfluous.
     
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