Geometric distribution probability of no success at all












1












$begingroup$


From S. Broverman, 2006:




A town's maintenance department has estimated that the cost of snow removal after a major
snowstorm is $100,000$. Historical information suggests that the number of major snowstorms in
a winter season follows a geometric distribution for which the probability of no major
snowstorms in a season is .4. The town purchases an insurance policy which pays nothing if
there is one or less major snowstorms in the season, but the insurance pays $50$% of all seasonal
snow removal costs for major snowstorms if there are $2$ or more major snowstorms. Find the
expected payout by the insurer.




I am confused by the bolded line "a geometric distribution for which the probability of no major
snowstorms in a season is $.4$". That sounds like it is saying the probability of no success at all is $.4$.



How does one find the probability of no success in a geometric distribution? From here it seems that one simply takes $(1-p)$ to the power of the $n$th attempt for which there is no success. However, according to that, how does one find the probability of no success at all, no matter how many attempts? Wouldn't that be $lim limits_{n to infty}(1-p)^n = 0$? If so, how would one understand the probability of no success being $.4$?






probability statistics







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  • 1




    $begingroup$
    As explained [here] the term "geometric distribution" is ambiguous, but I take it to mean that the probability of exactly $k$ snowstorms is $p(1-p)^k$ with $p=.4$
    $endgroup$
    – saulspatz
    Jan 14 at 17:00








  • 1




    $begingroup$
    Thank you, was there supposed to be link in the "[here]"?
    $endgroup$
    – agblt
    Jan 14 at 17:05






  • 1




    $begingroup$
    Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
    $endgroup$
    – saulspatz
    Jan 14 at 17:33
















1












$begingroup$


From S. Broverman, 2006:




A town's maintenance department has estimated that the cost of snow removal after a major
snowstorm is $100,000$. Historical information suggests that the number of major snowstorms in
a winter season follows a geometric distribution for which the probability of no major
snowstorms in a season is .4. The town purchases an insurance policy which pays nothing if
there is one or less major snowstorms in the season, but the insurance pays $50$% of all seasonal
snow removal costs for major snowstorms if there are $2$ or more major snowstorms. Find the
expected payout by the insurer.




I am confused by the bolded line "a geometric distribution for which the probability of no major
snowstorms in a season is $.4$". That sounds like it is saying the probability of no success at all is $.4$.



How does one find the probability of no success in a geometric distribution? From here it seems that one simply takes $(1-p)$ to the power of the $n$th attempt for which there is no success. However, according to that, how does one find the probability of no success at all, no matter how many attempts? Wouldn't that be $lim limits_{n to infty}(1-p)^n = 0$? If so, how would one understand the probability of no success being $.4$?






probability statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    As explained [here] the term "geometric distribution" is ambiguous, but I take it to mean that the probability of exactly $k$ snowstorms is $p(1-p)^k$ with $p=.4$
    $endgroup$
    – saulspatz
    Jan 14 at 17:00








  • 1




    $begingroup$
    Thank you, was there supposed to be link in the "[here]"?
    $endgroup$
    – agblt
    Jan 14 at 17:05






  • 1




    $begingroup$
    Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
    $endgroup$
    – saulspatz
    Jan 14 at 17:33














1












1








1





$begingroup$


From S. Broverman, 2006:




A town's maintenance department has estimated that the cost of snow removal after a major
snowstorm is $100,000$. Historical information suggests that the number of major snowstorms in
a winter season follows a geometric distribution for which the probability of no major
snowstorms in a season is .4. The town purchases an insurance policy which pays nothing if
there is one or less major snowstorms in the season, but the insurance pays $50$% of all seasonal
snow removal costs for major snowstorms if there are $2$ or more major snowstorms. Find the
expected payout by the insurer.




I am confused by the bolded line "a geometric distribution for which the probability of no major
snowstorms in a season is $.4$". That sounds like it is saying the probability of no success at all is $.4$.



How does one find the probability of no success in a geometric distribution? From here it seems that one simply takes $(1-p)$ to the power of the $n$th attempt for which there is no success. However, according to that, how does one find the probability of no success at all, no matter how many attempts? Wouldn't that be $lim limits_{n to infty}(1-p)^n = 0$? If so, how would one understand the probability of no success being $.4$?






probability statistics







share|cite|improve this question











$endgroup$




From S. Broverman, 2006:




A town's maintenance department has estimated that the cost of snow removal after a major
snowstorm is $100,000$. Historical information suggests that the number of major snowstorms in
a winter season follows a geometric distribution for which the probability of no major
snowstorms in a season is .4. The town purchases an insurance policy which pays nothing if
there is one or less major snowstorms in the season, but the insurance pays $50$% of all seasonal
snow removal costs for major snowstorms if there are $2$ or more major snowstorms. Find the
expected payout by the insurer.




I am confused by the bolded line "a geometric distribution for which the probability of no major
snowstorms in a season is $.4$". That sounds like it is saying the probability of no success at all is $.4$.



How does one find the probability of no success in a geometric distribution? From here it seems that one simply takes $(1-p)$ to the power of the $n$th attempt for which there is no success. However, according to that, how does one find the probability of no success at all, no matter how many attempts? Wouldn't that be $lim limits_{n to infty}(1-p)^n = 0$? If so, how would one understand the probability of no success being $.4$?






probability statistics




probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 16:54







agblt

















asked Jan 14 at 16:48









agbltagblt

350114




350114








  • 1




    $begingroup$
    As explained [here] the term "geometric distribution" is ambiguous, but I take it to mean that the probability of exactly $k$ snowstorms is $p(1-p)^k$ with $p=.4$
    $endgroup$
    – saulspatz
    Jan 14 at 17:00








  • 1




    $begingroup$
    Thank you, was there supposed to be link in the "[here]"?
    $endgroup$
    – agblt
    Jan 14 at 17:05






  • 1




    $begingroup$
    Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
    $endgroup$
    – saulspatz
    Jan 14 at 17:33














  • 1




    $begingroup$
    As explained [here] the term "geometric distribution" is ambiguous, but I take it to mean that the probability of exactly $k$ snowstorms is $p(1-p)^k$ with $p=.4$
    $endgroup$
    – saulspatz
    Jan 14 at 17:00








  • 1




    $begingroup$
    Thank you, was there supposed to be link in the "[here]"?
    $endgroup$
    – agblt
    Jan 14 at 17:05






  • 1




    $begingroup$
    Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
    $endgroup$
    – saulspatz
    Jan 14 at 17:33








1




1




$begingroup$
As explained [here] the term "geometric distribution" is ambiguous, but I take it to mean that the probability of exactly $k$ snowstorms is $p(1-p)^k$ with $p=.4$
$endgroup$
– saulspatz
Jan 14 at 17:00






$begingroup$
As explained [here] the term "geometric distribution" is ambiguous, but I take it to mean that the probability of exactly $k$ snowstorms is $p(1-p)^k$ with $p=.4$
$endgroup$
– saulspatz
Jan 14 at 17:00






1




1




$begingroup$
Thank you, was there supposed to be link in the "[here]"?
$endgroup$
– agblt
Jan 14 at 17:05




$begingroup$
Thank you, was there supposed to be link in the "[here]"?
$endgroup$
– agblt
Jan 14 at 17:05




1




1




$begingroup$
Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
$endgroup$
– saulspatz
Jan 14 at 17:33




$begingroup$
Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
$endgroup$
– saulspatz
Jan 14 at 17:33










1 Answer
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$begingroup$

From saulspatz's comment, I gather that the way to interpret the geometric distribution here is that success is "the point at which the snowstorms stop", and $p$ is the probability of the snowstorms stopping. Failure is that there was a snowstorm, therefore making it that the snowstorms did not yet stop.



Hence one can use the definition $p(1-p)^k$, and at $k=0$, this would mean that there was no failure because the snowstorms never started.



One can also use the definition $p(1-p)^{k-1}$ and at $k=1$, this would mean that on the first attempt, the snowstorm stopped, never starting.



See also my question here for an analogous case.






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    $begingroup$

    From saulspatz's comment, I gather that the way to interpret the geometric distribution here is that success is "the point at which the snowstorms stop", and $p$ is the probability of the snowstorms stopping. Failure is that there was a snowstorm, therefore making it that the snowstorms did not yet stop.



    Hence one can use the definition $p(1-p)^k$, and at $k=0$, this would mean that there was no failure because the snowstorms never started.



    One can also use the definition $p(1-p)^{k-1}$ and at $k=1$, this would mean that on the first attempt, the snowstorm stopped, never starting.



    See also my question here for an analogous case.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      From saulspatz's comment, I gather that the way to interpret the geometric distribution here is that success is "the point at which the snowstorms stop", and $p$ is the probability of the snowstorms stopping. Failure is that there was a snowstorm, therefore making it that the snowstorms did not yet stop.



      Hence one can use the definition $p(1-p)^k$, and at $k=0$, this would mean that there was no failure because the snowstorms never started.



      One can also use the definition $p(1-p)^{k-1}$ and at $k=1$, this would mean that on the first attempt, the snowstorm stopped, never starting.



      See also my question here for an analogous case.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        From saulspatz's comment, I gather that the way to interpret the geometric distribution here is that success is "the point at which the snowstorms stop", and $p$ is the probability of the snowstorms stopping. Failure is that there was a snowstorm, therefore making it that the snowstorms did not yet stop.



        Hence one can use the definition $p(1-p)^k$, and at $k=0$, this would mean that there was no failure because the snowstorms never started.



        One can also use the definition $p(1-p)^{k-1}$ and at $k=1$, this would mean that on the first attempt, the snowstorm stopped, never starting.



        See also my question here for an analogous case.






        share|cite|improve this answer











        $endgroup$



        From saulspatz's comment, I gather that the way to interpret the geometric distribution here is that success is "the point at which the snowstorms stop", and $p$ is the probability of the snowstorms stopping. Failure is that there was a snowstorm, therefore making it that the snowstorms did not yet stop.



        Hence one can use the definition $p(1-p)^k$, and at $k=0$, this would mean that there was no failure because the snowstorms never started.



        One can also use the definition $p(1-p)^{k-1}$ and at $k=1$, this would mean that on the first attempt, the snowstorm stopped, never starting.



        See also my question here for an analogous case.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 14 at 17:38

























        answered Jan 14 at 17:33









        agbltagblt

        350114




        350114






























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