Probability of inhabited planets: I have 16 exosystems, and each system is orbited by 3 planets....

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Bluebell

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Hello, can anyone help me with this question please? I'm flumoxed!

I have 16 exosystems, and each system is orbited by 3 planets. I know that 11 of the planets are inhabited by humanoids, but not sure of the distribution. However, I strongly suspect that one of the triple systems has both planets out of the three inhabited by humanoids. What is the probability of this?

Also what would the probability be if 2, or even 3 of the exosystems had both planets out of the three inhabited (still with a total of 11 planets inhabited)?

Thank you!
 
Bluebell said:
one of the triple systems has both planets out of the three inhabited
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What does "both" mean in this context? Does it mean exactly two? I'm not familiar with this usage.

In any case, please tell us a little about your context, so we can know how to answer. Is this from a math class, or a science fiction seminar, or what? And how much do you know about probability?

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Hi, yes, thanks for responding. In response to your question, yes, sorry, I meant exactly 2 out of 3 planets in a system, though it can be any 2. If no system has any 2 planets inhabited, then obviously any of 11 systems out of the 16 can have 1 planet inhabited. If one system has 2 planets inhabited, with 9 systems containing 1 inhabited planet, then would this probability simply be 10/16? If 2 systems have 2 planets inhabited with 7 systems containing 1 inhabited planet then is that probability 9/16? So would the probabilities simply be 11/16 or 10/16 and 9/16 respectively for each of the above 3 scenarios? It's a homework question, but is it just an easy trick question disguised in a science fiction scenario?!
 
Bluebell said:
Hi, yes, thanks for responding. In response to your question, yes, sorry, I meant exactly 2 out of 3 planets in a system, though it can be any 2. If no system has any 2 planets inhabited, then obviously any of 11 systems out of the 16 can have 1 planet inhabited. If one system has 2 planets inhabited, with 9 systems containing 1 inhabited planet, then would this probability simply be 10/16? If 2 systems have 2 planets inhabited with 7 systems containing 1 inhabited planet then is that probability 9/16? So would the probabilities simply be 11/16 or 10/16 and 9/16 respectively for each of the above 3 scenarios? It's a homework question, but is it just an easy trick question disguised in a science fiction scenario?!
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Now can you tell us what topics have been covered in your class, so we can try to help you use what you've learned, rather than something you've never heard of?

Also, please quote the problem exactly as given to you; I think you've paraphrased, which is very dangerous in probability questions! For example, are you sure it isn't that at least one system has at least two inhabited planets? That would be easier to solve, by considering the complement.
 
Hi, once again, thanks for responding, and yes, you're right, it does mention 'least' as you said! We've covered frequency distributions, standard deviations, ttests, chi squared tests, wilcoxon, mann-whitney, measures of dispersion, z scores, and combinations. The question was "There are 16 exosystems, each of which consists of three planets. Surveys show that 11 of the planets are inhabited by a humanoid species. However, the distribution of the 11 inhabited planets in the 16 systems is unknown. What is the likelihood that at least one system consists of two or more inhabited planets?"
 
Bluebell said:
Hi, once again, thanks for responding, and yes, you're right, it does mention 'least' as you said! We've covered frequency distributions, standard deviations, ttests, chi squared tests, wilcoxon, mann-whitney, measures of dispersion, z scores, and combinations. The question was "There are 16 exosystems, each of which consists of three planets. Surveys show that 11 of the planets are inhabited by a humanoid species. However, the distribution of the 11 inhabited planets in the 16 systems is unknown. What is the likelihood that at least one system consists of two or more inhabited planets?"
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Thanks.

Do you see how huge a difference those words you left out make? As I suggested, they will make this far easier than it would have been otherwise.

The topics you list are mostly statistics; this question is about probability. You'll need the basic definition of probability, and a little knowledge of combinations and permutations. And my hint about using the complement should have triggered some ideas, if you know enough about probability. Did it?

How many ways are there to choose 11 of the planets to be inhabited?

How many ways are there to choose 11 of the planets to be inhabited, if no two are in the same system?

Please show some attempt now.
 
Thank you for your reply! There are a total of 48 planets (3 X 16), so the number of combinations of ways that 11 planets might be distributed amongst 48 planets is 2.26 to the power of 10 (2.26E10). But the restriction is that 2 planets in a system are inhabited, so that would be 11 planets distributed amongst 16 systems which is 4368, but since there are 3 ways that 2 planets can be distributed in each triple system, this has to be multiplied by 3 to give 13104 combinations. So taking the fractionof the combinations by which 2 planets can be inhabited from a total of 11 in 16 triple planet systems is 13104/2.26E10 = 5.79E-7. Therefore the reciprocal of this is the probability which comes out as 1/5.79E-7 = 1 in 1,724,664?
 
Bluebell said:
Thank you for your reply! There are a total of 48 planets (3 X 16), so the number of combinations of ways that 11 planets might be distributed amongst 48 planets is 2.26 to the power of 10 (2.26E10). But the restriction is that 2 planets in a system are inhabited, so that would be 11 planets distributed amongst 16 systems which is 4368, but since there are 3 ways that 2 planets can be distributed in each triple system, this has to be multiplied by 3 to give 13104 combinations. So taking the fractionof the combinations by which 2 planets can be inhabited from a total of 11 in 16 triple planet systems is 13104/2.26E10 = 5.79E-7. Therefore the reciprocal of this is the probability which comes out as 1/5.79E-7 = 1 in 1,724,664?
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You're right that the number of possible ways to choose 11 inhabited planets is 48C11 = 2.26E10.

And if no two planets in a system are inhabited, we can choose 11 systems to have one planet inhabited in 16C11 = 4368 ways; you didn't say what you were doing very clearly there.

But we are not putting 2 planets in a system, so I don't know what you are saying next, and the rest falls apart.

Remember, the goal is to find the probability that no system has more than one inhabited planet, and then find the complement of that. We are not working with 2-inhabited planet systems.
 
Hi, thanks for your reply! I think I may have misinterpreted the question. Let me try to simplify my reasoning. I have 9 planets in 3 triple planet systems, and i want to know the likelihood of any 2 planets being inhabited from a random sampling of any 3 planets. There are 84 possible ways in which 3 planets could be chosen from 9 planets. So i call the systems A, B and C. The names of the planets in system A are 1, 2 and 3; in system B, 4, 5 and 6; in system C, 7, 8 and 9. So in system A i could have 1 and 2, 1 and 3, or 2 and 3 planets inhabited. In system B i could have 4 and 5, 4 and 6, or 5 and 6 inhabited. In system C i could have 7 and 8, 7 and 9, or 8 and 9 inhabited. Together these make 9 possible combinations of 2 planets being inhabited in the same system. So this comes out as 9/84 = 0.107. Taking the reciprocal, the chance that I will have 2 inhabited planets from the same system (either A, B or C), from taking a random sample of any 3 planets out of the 9, is 1 in 9.3, or approx 1 in 9. So if I take 9 samples of 3 random planets, one of them is highly likely to have 2 planets in it from system A, B or C...
 
Bluebell said:
Hi, thanks for your reply! I think I may have misinterpreted the question. Let me try to simplify my reasoning.

I have 9 planets in 3 triple planet systems, and i want to know the likelihood of any 2 planets being inhabited from a random sampling of any 3 planets. There are 84 possible ways in which 3 planets could be chosen from 9 planets. So i call the systems A, B and C. The names of the planets in system A are 1, 2 and 3; in system B, 4, 5 and 6; in system C, 7, 8 and 9.

So in system A i could have 1 and 2, 1 and 3, or 2 and 3 planets inhabited. In system B i could have 4 and 5, 4 and 6, or 5 and 6 inhabited. In system C i could have 7 and 8, 7 and 9, or 8 and 9 inhabited. Together these make 9 possible combinations of 2 planets being inhabited in the same system.

So this comes out as 9/84 = 0.107. Taking the reciprocal, the chance that I will have 2 inhabited planets from the same system (either A, B or C), from taking a random sample of any 3 planets out of the 9, is 1 in 9.3, or approx 1 in 9. So if I take 9 samples of 3 random planets, one of them is highly likely to have 2 planets in it from system A, B or C...
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The main problem is that you are now ignoring my advice entirely. If you'd like to try what I suggested, let me know. As it is, you're going back to something like your original attempt, which would be much harder.

But you seem to be doing an entirely different problem here, and you haven't quite stated what it is, so I can't judge it.
 
I am sorry, but that is the question I was asking - I have clarified this with others. Basically it is as stated. I just want to know what is the likelihood that if I'm looking at x triple planet exosystems, and if I take out a random sample of planets, y, then what is the chance that I would have any 2 planets inhabited out of the same triple system within the random sample (y). Anyway, thanks for your attempt at help, probably my fault for not explaining properly. Kind regards.
 
Bluebell said:
But the restriction is that 2 planets in a system are inhabited, so that would be 11 planets distributed amongst 16 systems which is 4368, but since there are 3 ways that 2 planets can be distributed in each triple system, this has to be multiplied by 3 to give 13104 combinations.
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You seem to be going in the right direction, but making mistakes along the way: 4368 is indeed the number of ways to distribute 11 planets among 16 systems, but this would come up in the case when there are no systems with more than one planet, i.e., the complimentary case.

Another hint: if there are 3 ways to pick a planet in each system what is the total number of ways to pick planets in those 11 systems?
 
Bluebell said:
I am sorry, but that is the question I was asking - I have clarified this with others. Basically it is as stated. I just want to know what is the likelihood that if I'm looking at x triple planet exosystems, and if I take out a random sample of planets, y, then what is the chance that I would have any 2 planets inhabited out of the same triple system within the random sample (y). Anyway, thanks for your attempt at help, probably my fault for not explaining properly. Kind regards.
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At first I thought this meant that when you quoted the actual problem, saying,
Bluebell said:
The question was "There are 16 exosystems, each of which consists of three planets. Surveys show that 11 of the planets are inhabited by a humanoid species. However, the distribution of the 11 inhabited planets in the 16 systems is unknown. What is the likelihood that at least one system consists of two or more inhabited planets?"
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you were ... lying.

After more thought, I think that what you are doing here is just restating it, emphasizing that the 11 planets are chosen randomly; saying that they are a random sample taken out, rather than a random sample that are inhabited, but then also describe which of those chosen are inhabited, is very confusing. This restatement didn't help me, though perhaps it helps you.

What you presumably mean is that you have 16 groups of 3, and randomly choose 11 of the 48 to call "inhabited"; and you want the probability that among those 11, at least two are in the same group.

I gave you a useful approach, and you sort of started that work, but then reverted to your original way, listing cases. We need to see you progress on the good way.

Do you understand the method we're working on? We're trying to find the probability that 11 random planets out of the 48, NONE are in the same system, which is the complement of what you want in the end. You found that there are 48C11 ways to choose the inhabited planets (which will be the denominator), and that there are 16C11 ways to choose 11 systems, each of which will have one planet. You need to take another step, finding the number of ways to choose one of the three planets in each of those 11 systems. Then you can put everything together.
 
Bluebell said:
So I guess that would be 16C11 - 11C3?
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Don't just guess ... state why you think so, and then decide whether the reason is convincing.

In combinatorics problems, I need a lot of convincing -- they can be tricky, so I don't believe my first thought!
 
There are 3 ways in which each of the 11 systems could have a single planet inhabited so this needs to be subtracted from all of the ways...
 
Bluebell said:
There are 3 ways in which each of the 11 systems could have a single planet inhabited so this needs to be subtracted from all of the ways...
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You really need to try and complete the problem, and, most importantly, not be afraid of making mistakes. You latest post is way to vague for anyone to react in a helpful way.
 
Bluebell said:
There are 3 ways in which each of the 11 systems could have a single planet inhabited so this needs to be subtracted from all of the ways...
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Think very, very, carefully -- which, as @blamocur says, doesn't mean not writing anything down, just writing and then thinking again!

I've picked 11 systems. For the first, I pick 1 of 3 planets; then, for the second, I pick 1 of 3 planets; and so on. How do I combine those numbers to get the total number of choices? Hint: you don't add or subtract them!
 
Hi @Bluebell,

(You were in the forum when I started writing this but I see you've now left; I trust you will come back again to get your problem solved? ?)

Reading through this thread I'm left wondering whether you have noted (& understood) a point that has come up more than once in the responses you've had.

Do you understand what the answers you've had mean when the they refer to the "complement"? It is not the "inverse" (ie: 1 over something or something raised to the power of ˉ1!)

I take it that you understand that a probability of 1 means that an event definitely will occur, while a probability of 0 means that an event definitely will not occur. This means that all probabilities lie in the range zero to one (inclusive).

Consider throwing a fair die. What is the probability of throwing a six?

Clearly it will be one in six or \(\displaystyle \scriptsize\frac{1}{6}\)

Now the complement of that is \(\displaystyle \scriptsize\frac{5}{6}\), ie: \(\displaystyle \large1-\scriptsize\frac{1}{6}\) (which is, in effect, the probability of everything else.).

So the probability of throwing a six \(\displaystyle \left(\scriptsize\frac{1}{6}\right)\) tells you how likely you are to throw a six (using a fair die) whilst its complement \(\displaystyle \left(\scriptsize\frac{5}{6}\right)\) tells you how unlikely you are to throw a six (or how likely you are to throw anything but a six, ie: any other of the numbers 1 to 5).

Does that explain how this question should be approached by finding the probability that none of the systems contain 2 or more inhabited planets (or that there are 11 of the systems that contain a single inhabited planet)? So then the complement of that probability (1 minus it) will be the probability that there are 2 or more inhabited planets in one (or more) of the systems (which is the answer you are looking for).

Hope that helps. ?
 
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