How many birds do I have to catch?

lonsdejlit

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Dear Forum,

I have the following problem with which I cannot solve. I have a very large population of birds e.g. 10 000. There are only 8 species of birds in this population. The size of each species is the same.

I would like to calculate how many birds I have to catch, to be sure in 80% that I caught one bird of each species.

I would be very obliged for help.
Lonsdejlit
 
Dear Forum,

I have the following problem with which I cannot solve. I have a very large population of birds e.g. 10 000. There are only 8 species of birds in this population. The size of each species is the same.

I would like to calculate how many birds I have to catch, to be sure in 80% that I caught one bird of each species.

I would be very obliged for help.
Lonsdejlit

Hi, welcome to the forum. :)

If you take a quick look at the rules, you'll see that we don't do your homework for you here, because you wouldn't learn any math that way.
https://www.freemathhelp.com/forum/announcement.php?f=35

But we are happy to help guide you through a problem, and provide help to get you unstuck from whatever you are stuck on. So to that end, could you please post what you done so far on this problem, or at least what thoughts you've had about how to go about solving it?

Am I right in thinking that "the size of each species is the same" means that the size of the population of each species is the same i.e. 10,000/8? If so, whenever you try to catch a bird, you're equally likely to catch one of each of the 8 species. So your probability of getting a bird of a given species on a given attempt would just be 1/8. That's a start, anyway.
 
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Thank you j-astron for the warm welcome and your willing to help. I've read the rules of the forum. I'm PhD candidate and it's not my homework. Nevertheless, maybe I paid too less attention to my homework that is why I'm struggling this issue. I need this calculation to estimate how many companies I have to survey to fill my case studies grid. I have 8 cases ;)

I tried the following approach:
What is the probability that I catch one, not having species in a first trial: 100%
What is the probability that I catch one new species when I couth already one: 7/8=87,5%
What is the probability that I catch one new species when I couth already two: 6/8=75%


I want to be confident in 80% of the results, so I multiply the probability by the confidence level getting:
100% * 80% = 0,8 (that is the amount of birds that I can catch with 0,8 probability)
87,5% * 80% = 0,9
75% * 80% = 1,1


Finally, sum of all the values comes to 17,8 (birds)
But it seems to be not the right solution...

Answering your question, yes you are right "the size of the population of each species is the same".

Kind regards,
Lonsdejlit
 
Thank you j-astron for the warm welcome and your willing to help. I've read the rules of the forum. I'm PhD candidate and it's not my homework. Nevertheless, maybe I paid too less attention to my homework that is why I'm struggling this issue. I need this calculation to estimate how many companies I have to survey to fill my case studies grid. I have 8 cases ;)

Ah ok. Well you've already managed to formulate a completely different problem that is mathematically-equivalent to yours, so I would say that you're already doing pretty well. By the way, in English, the past tense of "catch" is spelled "caught", not "couth".

I think that if we can solve this bird problem, it will tell you what you want to know, under the assumption that every company you survey is equally likely to fall under any one of your eight cases.

I'm sorry if I'm not understanding your attempt at a solution, but it seems like what you're computing is the probability that each subsequent capture attempt will yield a new species of bird. However, that is a different problem from the original one, which is what is the probability that you'll get at least one of every species of bird after n capture attempts?

Based on the original problem, I think we can ignore cases of < 8 capture attempts. You cannot get one of every species of bird if you capture fewer birds than the total number of species.

I think we can solve this problem by doing the counting explicitly.

Probability = (number of desired outcomes)/(number of possible outcomes)

= (number of combinations that include one of every species)/(total number of combinations)

So we need to compute both the numerator and denominator (top and bottom) of this ratio.

For the bottom (tallying all possible outcomes): you are choosing k items from a set of n = 8 items, with repetition allowed. The order in which you receive the items is irrelevant. So this is combination with repetition. Do you recall how to count combinations with repetition allowed?

For the top (tallying all desired outcomes): Right now I'm thinking it's

(number of ways of choosing one of every species) x (number of ways of choosing the remaining k - 8 birds)

= 1 x (8 choose k-8)

So for example, if k = 9, there is 1 way to select one bird of every species, so that we now have 8 birds. Then all we have to do is select who the remaining (9th) bird is. There is only one remaining bird to select because in this case (k - 8) = (9 - 8) = 1. There are (8 choose 1) = 8 ways to choose who the remaining bird is. So the number of desired outcomes is 1 x 8 = 8.

So basically, compute the probability starting at k = 8 and incrementing by 1, until you reach a k value at which the probability exceeds 0.8.
 
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Ah ok. Well you've already managed to formulate a completely different problem that is mathematically-equivalent to yours, so I would say that you're already doing pretty well. By the way, in English, the past tense of "catch" is spelled "caught", not "couth".

I think that if we can solve this bird problem, it will tell you what you want to know, under the assumption that every company you survey is equally likely to fall under any one of your eight cases.

I'm sorry if I'm not understanding your attempt at a solution, but it seems like what you're computing is the probability that each subsequent capture attempt will yield a new species of bird. However, that is a different problem from the original one, which is what is the probability that you'll get at least one of every species of bird after n capture attempts?

Based on the original problem, I think we can ignore cases of < 8 capture attempts. You cannot get one of every species of bird if you capture fewer birds than the total number of species.

I think we can solve this problem by doing the counting explicitly.

Probability = (number of desired outcomes)/(number of possible outcomes)

= (number of combinations that include one of every species)/(total number of combinations)

So we need to compute both the numerator and denominator (top and bottom) of this ratio.

For the bottom (tallying all possible outcomes): you are choosing k items from a set of n = 8 items, with repetition allowed. The order in which you receive the items is irrelevant. So this is combination with repetition. Do you recall how to count combinations with repetition allowed?

For the top (tallying all desired outcomes): Right now I'm thinking it's

(number of ways of choosing one of every species) x (number of ways of choosing the remaining k - 8 birds)

= 1 x (8 choose k-8)

So for example, if k = 9, there is 1 way to select one bird of every species, so that we now have 8 birds. Then all we have to do is select who the remaining (9th) bird is. There is only one remaining bird to select because in this case (k - 8) = (9 - 8) = 1. There are (8 choose 1) = 8 ways to choose who the remaining bird is. So the number of desired outcomes is 1 x 8 = 8.

So basically, compute the probability starting at k = 8 and incrementing by 1, until you reach a k value at which the probability exceeds 0.8.
I cannot check your proposal by computation since I cannot figure out what is the equation...
I'd appreciate if you could execute a computation
 
Actually I think I'm getting the counting wrong. I think I may have been wrong about the fact that ordering doesn't matter. E.g. if you survey three companies, and have 8 cases, and end up with (7,2,1), that is a different outcome from (2,1,7), and the possibility for different orderings also increases number of ways of getting all 8 cases (in the situation where you surveyed > 8 companies). So, apologies for the mistake.

If this is true, number of possible outcomes would be given by permutation with repetition. E.g. say I have n = 8 types of objects and I choose a set of k = 12 of them where the order matters. For each of the k buckets I have in which to put an object (e.g. traps for a bird), I have 8 possibilities, so I have (8*8*8*... *8*8) total possible outcomes, where the eights are repeated k times. Therefore I have n^k = 8^12 possible permutations in this case

The number of desired outcomes is tricky to compute the permutations for, and I have not figured it out yet. Perhaps other people will have ideas.

For what it's worth, I wrote code to simulate this using a random number generator, and I found that the fraction of outcomes having all 8 types of birds crossed over from 0.79 to 0.81 in going from 27 birds trapped to 28 birds trapped. So that suggests you need to survey at least 28. But I haven't figured out how to show that, sorry.
 
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