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Thread: Geometrical distribution problem

  1. #1
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    Geometrical distribution problem

    I am trying to solve this problem:

    X_1, X_2,..., X_100 are independent and have geometrical distribition with probability of success 1/2

    1. P[X_1>10]

    I'm not sure what am I computing here... Is it the probability that success will occur in first event after 10 failures?

    2. P[X_1+X_2>3]

    Is this the probability of success occuring in first or second event after at least 3 failures?

    4. P[2*X_1+X_2>3]

    I have no idea how to interpret this.

    5. E[X_1+X_2]

    E[X_1+X_2] = E[X_1] + E[X_2] = 1*1/2 + 1*1/2 = 1
    Is this correct?
    Last edited by bessi; 05-14-2014 at 04:19 PM. Reason: tags

  2. #2
    Points: 2,513, Level: 30
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    Re: Geometrical distribution problem

    1. Yes.

    2. No. Read what it says; X1+X2 is the sum of two independent random variables. You are finding the probability that first success for X1 plus X2 is greater than 3.

    4. Same as above.

    5. No. I'm sure you have notes on the mean of a geometric random variable. Those "individual" means don't make any sense whatsoever, which should be the first clue that something is wrong.

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