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Flipping coin

  1. Mar 25, 2013 #1
    flipping a coin 6 times and let K be the number of crowns that we have after flipping the coin 6 times. The coin has a probability of 0.3 to get the crown.

    P(K=3)=?

    if we flip the coin 6 times we will have 31 crowns but how i can estimate this series of crowns?
     
  2. jcsd
  3. Mar 25, 2013 #2

    MathematicalPhysicist

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    Gold Member

    To achieve three here's what you do:

    the easy way is to calaulate when we don't achieve three crowns and then subtract 1 from it.

    We might not have three crowns if:
    1. there was none. which is 0.7^6
    2. there was just one crown which is: [tex]{6\choose 1} *0.3*0.7^5[/tex]
    3. there were only two crowns which is [tex] {6\choose 2} *0.3^2 * 0.7^4[/tex]
     
  4. Mar 25, 2013 #3
    i will sum this? how i can estimate 2 and 3 with nCk ?
     
  5. Mar 25, 2013 #4
    i did it but its says that the result its not correct ....
     
  6. Mar 25, 2013 #5
    I think you mean "...we will have 3 crowns...".

    You will have 3 crowns after flipping a coin 6 times in (6 choose 3) different ways, right? And you get a crown with probability p = 0.3, which implies that you will not have a crown with probability q = 1-p = 0.7. Suppose one of the experiments results as HHTTTH (which is one of the (6 choose 3) possible ways to get 3 crowns). The probability of getting this result of the experiment is

    [tex]0.3^3\times 0.7^3[/tex]

    You have to sum these probabilities for all (6 choose 3) ways.

    Your question is just an example for the binomial distribution.
     
  7. Mar 25, 2013 #6
    its not correct.....
     
  8. Mar 25, 2013 #7
    Your answer finds P(K>=3), but OP asks for P(K=3).
     
  9. Mar 25, 2013 #8
    Can you give your reasoning? May I ask what your computational result is?
     
  10. Mar 25, 2013 #9
    i found it finaly thanks for help man.
     
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