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Math Help - help with geometric probability

  1. #1
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    help with geometric probability

    hey

    I'm stuck on one of the question in our online test in statistics:


    Suppose that X ε [1,2,..., ] is discrete random variable that follows a geometric distribution with mean 3.74.
    What is the probability that X is an even number?

    Give your solution accurate to 4 decimal places.

    my idea was:

    from geometric sequence:

    x=1/theta

    so:

    3.14=1/theta -> theta=0.26737968

    and then use P(X=x)=(1-theta)^x-1 * theta

    to get P=0.1373945

    however don't know if that's what the question is asking about (i.e what is the 'successs' in this geometric distribution, number being even?) another idea was to put:

    P(X=2x), if that makes any sense, to account for the number being even however don't know how to go from there

    any thoughts?
    would appreciate any help
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  2. #2
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    Re: help with geometric probability

    Quote Originally Posted by bart123456 View Post
    hey

    I'm stuck on one of the question in our online test in statistics:


    Suppose that X ε [1,2,..., ] is discrete random variable that follows a geometric distribution with mean 3.74.
    What is the probability that X is an even number?

    Give your solution accurate to 4 decimal places.

    my idea was:

    from geometric sequence:

    x=1/theta

    so:

    3.14=1/theta -> theta=0.26737968

    and then use P(X=x)=(1-theta)^x-1 * theta

    to get P=0.1373945

    however don't know if that's what the question is asking about (i.e what is the 'successs' in this geometric distribution, number being even?) another idea was to put:

    P(X=2x), if that makes any sense, to account for the number being even however don't know how to go from there

    any thoughts?
    would appreciate any help
    Mean = 1/p => p = 1/3.75 = 4/15.

    Pr(X = even) = Pr(X = 2) + Pr(X = 4) + Pr(X = 6) + .... = 1 - p using the pmf of the geometric distribution and the formula for the sum of an infinite series.

    Details are left for you to do.
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  3. #3
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    Re: help with geometric probability

    Thank you so much!
    Maybe I don't understand subject so well, but we only did infinite series sums if
    -1<r<1 so don't know how to compute the sum with r=2.

    I guessed to use formula

    ((1-θ)^2i-1)θ

    and sum it from i=1 up to infinity
    but I don't know how to obtain a numerical answer not in terms of ratio of i.
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  4. #4
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    Re: help with geometric probability

    Quote Originally Posted by bart123456 View Post
    Thank you so much!
    Maybe I don't understand subject so well, but we only did infinite series sums if
    -1<r<1 so don't know how to compute the sum with r=2.

    I guessed to use formula

    ((1-θ)^2i-1)θ

    and sum it from i=1 up to infinity
    but I don't know how to obtain a numerical answer not in terms of ratio of i.
    The common ratio r is NOT 2. I don't know how you got that since you haven't shown your working. The common ratio is a number between 0 and 1 (otherwise the infinite series would not sum to a finite number!).
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