Geometric distribution probability of no success at all
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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
asked Jan 14 at 16:48
agbltagblt
350114
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add a comment |
$begingroup$
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
asked Jan 14 at 16:48
agbltagblt
350114
$endgroup$
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
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Thank you, was there supposed to be link in the "[here]"?
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– agblt
Jan 14 at 17:05
1
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Yes there was: en.wikipedia.org/wiki/Geometric_distribution Sorry
$endgroup$
– saulspatz
Jan 14 at 17:33
add a comment |
$begingroup$
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
asked Jan 14 at 16:48
agbltagblt
350114
$endgroup$
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
asked Jan 14 at 16:48
agbltagblt
350114
asked Jan 14 at 16:48
agbltagblt
350114
edited Jan 14 at 16:54
agblt
asked Jan 14 at 16:48
agbltagblt
350114
asked Jan 14 at 16:48
agbltagblt
350114
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Jan 14 at 17:33
agbltagblt
350114
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add a comment |
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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.
answered Jan 14 at 17:33
agbltagblt
350114
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 14 at 17:33
agbltagblt
350114
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 14 at 17:33
agbltagblt
350114
$endgroup$
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.
answered Jan 14 at 17:33
agbltagblt
350114
edited Jan 14 at 17:38
answered Jan 14 at 17:33
agbltagblt
350114
answered Jan 14 at 17:33
agbltagblt
350114
answered Jan 14 at 17:33
agbltagblt
350114
350114
add a comment |
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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