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Amazingly simple probability question

  1. Feb 21, 2005 #1
    I know this is a really easy problem, but i am confusing myself with it...

    there are three people who each have two possible actions-pay, or not pay (with symmetric probabilities of p for pay, and 1-p for not pay).

    what is the probability that at least two people pay?

    I first thought that it was just p^2, but that can't be right because that doesn't take into account the third person... so I then thought that it was 3p^2, because there are three possible combinations of 2 people paying. Is this all I need to do?

    thanks

    ~feeling stupid
     
  2. jcsd
  3. Feb 21, 2005 #2
    Just remember binomial probability. The probability that @least 2people pay means either 2 or 3 people pay. Thus, apply the binomial probability formula:
    (where p=probability of paying)

    [3!/(2!*1!)]*(p^2)*(1-p) + [3!/(3!*0!)]*(p^3)
    (As you can see, I don't know LaTex)

    If the probability of paying is 50%, then the above expression should amount to 0.5,
    meaning that there is a 50%chance that @least two people will pay.
     
  4. Feb 21, 2005 #3

    arildno

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    Just elaborating on bomba's reply:
    [tex]1=1^{3}=((1-p)+p)^{3}=\sum_{i=0}^{3}\binom{3}{i}(1-p)^{i}p^{3-i}[/tex]
    Hence, your probability is:
    [tex]\sum_{i=0}^{1}\binom{3}{i}(1-p)^{i}p^{3-i}[/tex]
     
  5. Feb 24, 2005 #4
    Hey, can you teach me LaTex? :redface: That'd be great :rofl:
     
  6. Feb 24, 2005 #5

    arildno

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    Click on the images to see how the code is written.
     
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