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Statistics-Probability Distribution of Discrete Random Variable

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent is independent of previous encounters. The player continues to contest opponents until defeated.

    What is the probability that a player defeats at least two opponents in a game?
    What is the probability that a player contests four or more opponents in a game?
    What is the expected number of game plays until a player contests four or more opponents?

    2. Relevant equations

    f(x)=(1-p)^(x-1)*p
    E=1/p

    3. The attempt at a solution

    I know that these are Bernoulli trials.
    I chose Geometric distribution because the number of 'trials' is not fixed.

    Defeat at least two opponents:
    pmf(1)+pmf(2)=cmf(2)=0.2+0.13=0.36

    Contest four or more:
    1-cmf(3)=1-[pmf(3)+pmf(2)+pmf(1)]=1-0.488=0.512

    Expected games to contest four or more:

    1/0.512=1.9???This is an illogical answer.
     
  2. jcsd
  3. Sep 15, 2012 #2

    LCKurtz

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    You might start by stating what your random variable ##X## represents. Although I can guess, you should tell us because there are a couple of ways the geometric distribution is set up and it needs to be clear to both of us.
     
  4. Sep 15, 2012 #3
    X represents the number of opponents faced per set of trials.
     
  5. Sep 16, 2012 #4

    LCKurtz

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    You haven't shown enough calculation to follow what you did. I suspect you may be using the wrong formula for ##f(x)##. Are you using ##f(x) = .8^{x-1}\cdot 2##? So to defeat at least 2 opponents you want ## P(X \ge 3) = 1 - P(X=1)-P(x=2)##. I get ##.64## for that one.
     
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