Chance of finding a tree in a sample area
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If I have 83 trees appearing over 24 quadrats (a quadrat being a 10m x 10m sample area => 2400m^2 total area), what's the chance of finding a tree in a sample area? What's the chance of finding a tree per m^2?
probability statistics
asked Nov 23 at 2:32
John Smith
82
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If I have 83 trees appearing over 24 quadrats (a quadrat being a 10m x 10m sample area => 2400m^2 total area), what's the chance of finding a tree in a sample area? What's the chance of finding a tree per m^2?
probability statistics
asked Nov 23 at 2:32
John Smith
82
What information do you have about the distribution of the trees?
– angryavian
Nov 23 at 2:37
Nothing beyond what's stated in the question; 83 trees appear over 24 quadrats. I need to find the frequency, quoted as "chance of finding a species in the sample area". Example given to me is: 10 occurences in 20 quadrats => a frequency of 50%. From which I would assume the frequency is 346%, if not for the "chance of finding" in the definition...
– John Smith
Nov 23 at 2:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I have 83 trees appearing over 24 quadrats (a quadrat being a 10m x 10m sample area => 2400m^2 total area), what's the chance of finding a tree in a sample area? What's the chance of finding a tree per m^2?
probability statistics
asked Nov 23 at 2:32
John Smith
82
If I have 83 trees appearing over 24 quadrats (a quadrat being a 10m x 10m sample area => 2400m^2 total area), what's the chance of finding a tree in a sample area? What's the chance of finding a tree per m^2?
probability statistics
probability statistics
asked Nov 23 at 2:32
John Smith
82
asked Nov 23 at 2:32
John Smith
82
edited Nov 23 at 2:54
asked Nov 23 at 2:32
John Smith
82
asked Nov 23 at 2:32
John Smith
82
asked Nov 23 at 2:32
John Smith
82
82
What information do you have about the distribution of the trees?
– angryavian
Nov 23 at 2:37
Nothing beyond what's stated in the question; 83 trees appear over 24 quadrats. I need to find the frequency, quoted as "chance of finding a species in the sample area". Example given to me is: 10 occurences in 20 quadrats => a frequency of 50%. From which I would assume the frequency is 346%, if not for the "chance of finding" in the definition...
– John Smith
Nov 23 at 2:53
add a comment |
What information do you have about the distribution of the trees?
– angryavian
Nov 23 at 2:37
Nothing beyond what's stated in the question; 83 trees appear over 24 quadrats. I need to find the frequency, quoted as "chance of finding a species in the sample area". Example given to me is: 10 occurences in 20 quadrats => a frequency of 50%. From which I would assume the frequency is 346%, if not for the "chance of finding" in the definition...
– John Smith
Nov 23 at 2:53
What information do you have about the distribution of the trees?
– angryavian
Nov 23 at 2:37
What information do you have about the distribution of the trees?
– angryavian
Nov 23 at 2:37
Nothing beyond what's stated in the question; 83 trees appear over 24 quadrats. I need to find the frequency, quoted as "chance of finding a species in the sample area". Example given to me is: 10 occurences in 20 quadrats => a frequency of 50%. From which I would assume the frequency is 346%, if not for the "chance of finding" in the definition...
– John Smith
Nov 23 at 2:53
Nothing beyond what's stated in the question; 83 trees appear over 24 quadrats. I need to find the frequency, quoted as "chance of finding a species in the sample area". Example given to me is: 10 occurences in 20 quadrats => a frequency of 50%. From which I would assume the frequency is 346%, if not for the "chance of finding" in the definition...
– John Smith
Nov 23 at 2:53
add a comment |
1 Answer
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Given no more information, the chance of a particular tree being in a given quadrat is $frac 1{24}$. Therefore the chance of the tree not being in the quadrat is $frac {23}{24}$. Lacking any other information, we assume that the placement of the trees is independent (a dubious assumption, since trees tend to cluster because of how they propagate). Therefore the chance of our quadrat not having any of the 83 trees would be $left(frac {23}{24}right)^{83}$.
And thus the probability that our quadrat does have a tree in it is $$1 -left(frac {23}{24}right)^{83}$$
The calculation for sample sizes of $1; m^2$ is similar.
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Given no more information, the chance of a particular tree being in a given quadrat is $frac 1{24}$. Therefore the chance of the tree not being in the quadrat is $frac {23}{24}$. Lacking any other information, we assume that the placement of the trees is independent (a dubious assumption, since trees tend to cluster because of how they propagate). Therefore the chance of our quadrat not having any of the 83 trees would be $left(frac {23}{24}right)^{83}$.
And thus the probability that our quadrat does have a tree in it is $$1 -left(frac {23}{24}right)^{83}$$
The calculation for sample sizes of $1; m^2$ is similar.
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
add a comment |
up vote
0
down vote
accepted
Given no more information, the chance of a particular tree being in a given quadrat is $frac 1{24}$. Therefore the chance of the tree not being in the quadrat is $frac {23}{24}$. Lacking any other information, we assume that the placement of the trees is independent (a dubious assumption, since trees tend to cluster because of how they propagate). Therefore the chance of our quadrat not having any of the 83 trees would be $left(frac {23}{24}right)^{83}$.
And thus the probability that our quadrat does have a tree in it is $$1 -left(frac {23}{24}right)^{83}$$
The calculation for sample sizes of $1; m^2$ is similar.
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Given no more information, the chance of a particular tree being in a given quadrat is $frac 1{24}$. Therefore the chance of the tree not being in the quadrat is $frac {23}{24}$. Lacking any other information, we assume that the placement of the trees is independent (a dubious assumption, since trees tend to cluster because of how they propagate). Therefore the chance of our quadrat not having any of the 83 trees would be $left(frac {23}{24}right)^{83}$.
And thus the probability that our quadrat does have a tree in it is $$1 -left(frac {23}{24}right)^{83}$$
The calculation for sample sizes of $1; m^2$ is similar.
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
Given no more information, the chance of a particular tree being in a given quadrat is $frac 1{24}$. Therefore the chance of the tree not being in the quadrat is $frac {23}{24}$. Lacking any other information, we assume that the placement of the trees is independent (a dubious assumption, since trees tend to cluster because of how they propagate). Therefore the chance of our quadrat not having any of the 83 trees would be $left(frac {23}{24}right)^{83}$.
And thus the probability that our quadrat does have a tree in it is $$1 -left(frac {23}{24}right)^{83}$$
The calculation for sample sizes of $1; m^2$ is similar.
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
answered Nov 23 at 14:17
Paul Sinclair
19.1k21441
19.1k21441
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
add a comment |
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Thanks for the response! While I don't need the answer, would I, therefore, be correct to calculate the probability 1 m^2 does contain a tree through 1 - (2399/2400)^83? Also, so I can read up on it, is there a name for the general form of the formula?
– John Smith
Nov 23 at 16:53
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
Yes, that is the probability for the $m^2$ problem. I know of no name. It is just a general principle of probability that sometimes it is easier to directly calculate $P(lnot A)$ than it is to calculate $P(A)$. Because $A$ and $lnot A$ are mutually exclusive and include all possibilities, $P(A) + P(lnot A) = 1$. So if you can calculate one, you can get the other by subtraction.
– Paul Sinclair
Nov 24 at 17:45
add a comment |
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What information do you have about the distribution of the trees?
– angryavian
Nov 23 at 2:37
Nothing beyond what's stated in the question; 83 trees appear over 24 quadrats. I need to find the frequency, quoted as "chance of finding a species in the sample area". Example given to me is: 10 occurences in 20 quadrats => a frequency of 50%. From which I would assume the frequency is 346%, if not for the "chance of finding" in the definition...
– John Smith
Nov 23 at 2:53