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Math Help - Geometric Distributions - Problem

  1. #1
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    Angry Geometric Distributions - Problem

    Quote: A Concise Course in Advanced Statistics - Crawshaw, Pg. 277 Question 10.


    X~Geo(p) and the probability that the first success is obtained on the second attempt is 0.1275.

    If p > 0.5, find P(X > 2)
    .


    Textbook ANS: 0.7225


    Reminding, Geometric Model:
    P(X=r) = q^{r-1}p, r = 1, 2, 3..\infty. And to the limit \sum_{r=1 }^{\infty }P(X=r)=1
    Where,
    r : rth attempt
    q : the probability of failure
    p : the probability of success


    p (the probability of success) is a constant, but I have trouble understanding what p > 0.5 really means from the quoted question above.

    Does it means a range of values of p then how to solve the question,
    or it just simply requires one constant p = 0.6.

    If that is so, there I am trying to solve it as follows:

    q^{2-1}p=0.1275

    P(>2)=1-P(X\leq 2)

    Therefore,

    1-P(X\leq 2)=1-\left [ P(1) +P(2)\right ]=1-\left [ p+qp \right ]=1-\left [ {\color{Red} 0.6}+0.1275 \right ] = 0.2725 \neq 0.7225


    You are welcome to give your help. Thanks.
    Last edited by zikcau25; March 20th 2014 at 05:23 AM.
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  2. #2
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    Re: Geometric Distributions - Problem

    In order that "the first success occurs on the second trial", there must be failure on the first trial and a success on the second. The probability of failure on the first attempt is 1- p and the probability of success on the second trial is (1- p)p. So the probability of "first success on the second trial is (1- p)^2p= 0.1275

    Solve p- p^3= 0.1275 or p^3- p- 0.1275. You don't say how you got p= 0.6. It certainly does not satisfy that. I get between 9.0 and 9.1.
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  3. #3
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    Re: Geometric Distributions - Problem

    Quote Originally Posted by zikcau25 View Post
    X~Geo(p) and the probability that the first success is obtained on the second attempt is 0.1275.
    If p > 0.5, find P(X > 2).
    Textbook ANS: 0.7225
    I think that the given answer should be $0.0225$. It is just a typo.

    $\mathcal{P}(X=2)=(1-p)p=0.1275$ gives $p=0.85$ (that is p>.5).

    So $1-(0.85+0.1275)=0.0225$.
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  4. #4
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    Cool Re: Geometric Distributions - Problem

    Thanks to Sir HallsofIvy, for reminding me to evaluate, quadratically, two values of p from the relationship p = (1- q) to choose only one value satisfying P > 0.5.

    And Secondly, to Sir Plato for confirming the textbook printing error for P > 0.5 instead of P < 0.5
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