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

[Bluebell]

Thank you Highlander. I think I'm getting mixed up trying to think about complementation and probability together. I have 16 triple planet systems. I find that I have 11 inhabited planets. I don't know whether the 11 planets are from 1 system each, or whether some of the systems have more than 1 inhabited planet. But I just want to work out the probability of the latter scenario. Therefore I first need to work out the first part - that the 11 planets are from 1 system each, and by subtracting that probability from 1, it will give me the probability of the second scenario. In a nutshell, is that the approach?

So first I work out the probability that 11 inhabited planets belong to individual systems. But there are 16 systems, from which any combination of 11 planets could be chosen. So first I work out how many combinations that could represent (16C11). Then imaginarily, each of those possible combinations of triple systems has 1 inhabited planet in it. So what is the probability of that? So it is a third per system. So then the probability would be a third times by 16C11. So the probability of more than one planet being inhabited is 1 minus a third multiplied by itself 16C11 times, which makes it highly probable that the 11 inhabited planets woud come from more than one planet per system.

 

[The Highlander]

So first I work out the probability that 11 inhabited planets belong to individual systems. But there are 16 systems, from which any combination of 11 planets could be chosen. So first I work out how many combinations that could represent (16C11). Then imaginarily, each of those possible combinations of triple systems has 1 inhabited planet in it. So what is the probability of that? So it is a third per system. So then the probability would be a third times by 16C11. So the probability of more than one planet being inhabited is 1 minus a third multiplied by itself 16C11 times, which makes it highly probable that the 11 inhabited planets woud come from more than one planet per system.

Now you're getting there (almost)! ?

The Question: "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?"

If none of the systems have more than 1 inhabited planet then there must be 11 systems with an inhabited planet. If any system were to have 2 (or more) inhabited planets then there would have to be 10 (or fewer) systems that had such planets. Yes?

Therefore, you simply want to know the probability that there are 11 systems with an inhabited planet.

So how many ways could you choose 11 systems out of 16? That's already been "agreed" as \(\displaystyle _{16}C_{11}\). Yes?

So what is the probability that there will be 11 systems with an inhabited planet?

And then what will the probability be that that is not the case (ie: the complement)?

The fact that you are told that each system contains three planets is a complete red herring (which you just need to ignore). It wouldn't matter how many planets each system contained (from zero to whatever)! If there are only 11 inhabited planets then they must be spread across 11 systems for there to be no system with two or more; if any system contained 2 or more inhabited planets then there would be fewer than 11 systems containing them!

So once you know the probability that there are 11 systems containing an inhabited planet, it's a simple matter to find the probability that that is not the case, ie: that at least one system contains 2 or more inhabited planets (which, again, is exactly what the question asks.)

It's really quite simple (when you think about it) isn't it?

Now go finish it off and tell us your (final) answer. ?

PS: The "others" that you "clarified" this with were clearly just as confused as you were. Tell us the right answer and you can then go and sort them out! ?
(Feel free to quote anything that's been provided in this thread for you; you'll be the class Genius! ?)

Cheers. ?

 

[Dr.Peterson]

Thank you Highlander. I think I'm getting mixed up trying to think about complementation and probability together. I have 16 triple planet systems. I find that I have 11 inhabited planets. I don't know whether the 11 planets are from 1 system each, or whether some of the systems have more than 1 inhabited planet. But I just want to work out the probability of the latter scenario. Therefore I first need to work out the first part - that the 11 planets are from 1 system each, and by subtracting that probability from 1, it will give me the probability of the second scenario. In a nutshell, is that the approach?

So first I work out the probability that 11 inhabited planets belong to individual systems. But there are 16 systems, from which any combination of 11 planets could be chosen. So first I work out how many combinations that could represent (16C11). Then imaginarily, each of those possible combinations of triple systems has 1 inhabited planet in it. So what is the probability of that? So it is a third per system. So then the probability would be a third times by 16C11. So the probability of more than one planet being inhabited is 1 minus a third multiplied by itself 16C11 times, which makes it highly probable that the 11 inhabited planets woud come from more than one planet per system.

Thanks for writing out details of your understanding; that is important.

The only error here is what I bolded.

You are choosing one of three for each of the inhabited systems, which means 11 times, not 16C11 times. Do you see this? And you are doing this after first choosing the systems, which can be done in 16C11 ways.

The approach to probability that I have been suggesting is to put the total number of ways to choose inhabited planets in the denominator, and the number of ways that fit the requirement (that there is only one inhabited per system) in the numerator. What you have said is more or less equivalent, but in my mind a little harder to be sure of.

The fact that you are told that each system contains three planets is a complete red herring (which you just need to ignore). It wouldn't matter how many planets each system contained (from zero to whatever)! If there are only 11 inhabited planets then they must be spread across 11 systems for there to be no system with two or more; if any system contained 2 or more inhabited planets then there would be fewer than 11 systems containing them!

I think you are wrong here. The 3 planets per system does affect the answer.

 

[blamocur]

The fact that you are told that each system contains three planets is a complete red herring (which you just need to ignore). It wouldn't matter how many planets each system contained (from zero to whatever)!

I strongly disagree with this statement. You have 16C11 ways to distribute 11 inhabited planets among 16 systems, but then you can pick a different planet in each of those 11 chosen systems.

 

[blamocur]

You are choosing one of three for each of the inhabited systems, which means 11 times

I am guessing you mean 3 times, not 11.

 

[The Highlander]

I think you are wrong here. The 3 planets per system does affect the answer.

Can you explain how, please? (In a private conversation if you feel you don't want to "give away" anything.)
I don't see how it matters whether an inhabited planet is the first, second or third one orbiting each system's star!
I think the logic holds that if no system has more than 1 of the (available) 11 inhabited planets then there must be a total of 11 systems containing the inhabited planets. So \(\displaystyle _{16}C_{11}\) gets you number of ways such planets could be distributed among the 16 systems and, thence, the probability that no system contains more than 1 inhabited planet, surely?

 

[The Highlander]

I strongly disagree with this statement. You have 16C11 ways to distribute 11 inhabited planets among 16 systems, but then you can pick a different planet in each of those 11 chosen systems.

Why does it matter (a jot) which of the three planets in any one system is the inhabited one?
If there is any system that contains 2 or more inhabited planets then the number of systems that contain any inhabited planets must be reduced below 11!

 

[blamocur]

Why does it matter (a jot) which of the three planets in any one system is the inhabited one?
If there is any system that contains 2 or more inhabited planets then the number of systems that contain any inhabited planets must be reduced below 11!

When you consider the complimentary case where there is at most 1 inhabited planet in any system you have 16C11 ways to pick which systems, but within each of those picked systems you have 3 ways to pick a particular planet. They are different cases if only because you counted them as different in the total of 48C11 cases.

 

[blamocur]

I don't see how it matters whether an inhabited planet is the first, second or third one orbiting each system's star!

I am repeating myself here, but still: you do consider them as different cases when you count the total of 48C11 choices.

 

[Dr.Peterson]

You are choosing one of three for each of the inhabited systems, which means 11 times, not 16C11 times.

I am guessing you mean 3 times, not 11.

No, I meant you are making 11 choices, each time one out of 3.

I don't see how it matters whether an inhabited planet is the first, second or third one orbiting each system's star!
I think the logic holds that if no system has more than 1 of the (available) 11 inhabited planets then there must be a total of 11 systems containing the inhabited planets. So 16C11 gets you number of ways such planets could be distributed among the 16 systems and, thence, the probability that no system contains more than 1 inhabited planet, surely?

The denominator of the probability is the total number of ways to choose individual planets. So the numerator must also distinguish individual planets. Just selecting 11 systems doesn't do so.

 

[The Highlander]

Hi (again) @Bluebell, ?

Have you paid attention to everything that's been said after my last Reply to you?

It seems I may have oversimplified the situation and you can't just ignore the number of planets in each system! ?‍

My esteemed colleagues (they do know much better than me) have persuaded me that you must take into account the possible configurations of the planets in each (of the 11) systems that might contain an inhabitable one in order to account for all the possible outcomes where no system contains more than one inhabitable planet (as opposed to all the possible combinations of all 48 planets).

Once you have that figured out you should be good to go in finding the probability of that occurring and, thence, the (complementary) probability that one (or more) system(s) might contain two or more inhabitable planets.

If you're (still) stuck on that, just come back and let us know; further help will then, I'm sure, be offered.

Please do let us know what final answer you arrive at. ?

 

[Bluebell]

Hi guys, thanks for all your replies. I think I get it. So I know that I have 11 inhabited planets. Of the 48 total planets, I haven't a clue where the 11 inhabited planets might be, so the number of possibilities are 48C11 which is 2.26E10. But from what Dr. Peterson was saying, then I need to work out the number of combinations for just one planet in eachof the 16 systems being inhabited. So I can represent each triple system as 1 to 16. It doesn't matter which planet in the 16 triple systems is inhabited, it is just one of them, and there are 16C11 ways for getting 11 singly inhabited planets out of 16 systems. So then, as a fraction of the whole this is 16C11/2.26E10. So that gives me the fraction of the 2.26E10 combinations for which only 1 planet could be inhabited from all the possible combinations. But since I want to know the fraction for more than 1 planet, it is 1 - (16C11/2.26E10), which comes out as 0.999999807. But is that fraction a probability, or just a fraction?

But maybe the above doesn't matter because I just want to know the probability of 1 planet being inhabited from 11 systems, no matter whatever the combination of 11 from 48 planets is. Then that would just be a third multiplied by itself 11 times which is 5.64E-6. So subtracting that from one to give the number of planets that have more than one inhabitant is 0999994355, which I think might be what Dr Peterson was getting at?

I think I'm going to go lie down now, my brain is starting to hurt... ;-)

 

[blamocur]

I need to work out the number of combinations for just one planet in eachof the 16 systems being inhabited.

Not more than one planet. I.e., one or zero planets.

 

[Dr.Peterson]

But from what Dr. Peterson was saying, then I need to work out the number of combinations for just one planet in each of the 16 systems being inhabited.

You mean, for each of the 11 planets that are inhabited, right?

Then that would just be a third multiplied by itself 11 times which is 5.64E-6.

Not quite. As I would take it, [imath]3^{11}=177147[/imath], which will be part of your numerator. (See below.)

I'll repeat what I said before:

The approach to probability that I have been suggesting is to put the total number of ways to choose inhabited planets in the denominator, and the number of ways that fit the requirement (that there is only one inhabited per system) in the numerator. What you have said is more or less equivalent, but in my mind a little harder to be sure of.

Don't start talking about probability too soon. Find the denominator, which you have, and then the numerator, which is the number of ways to (a) choose 11 of the 16 systems, and (b) then choose 1 of the 3 planets in each of those 11 systems. THEN put the numerator over the denominator, and subtract from 1.

 

[Dr.Peterson]

You mean, for each of the 11 planets that are inhabited, right?

I intended to say, as I hope is clear from the last paragraph,

You mean, for each of the 11 systems that are inhabited, right?​

 

[Bluebell]

Ok, thanks. For each of the 11 systems, not planets. I see. So I have 16C11 for the number of possible ways I could chose 11 inhabited planets from 16 systems which is 4368, but that still doesn't tell me whether it is only 1 or more than 1 planet inhabited for each of those 11 systems. So now I have to somehow get to 1 inhabited planet for all of the combinations of 11 systems which is the (b) part of your comment. So 3^11 gives the total number which is 177147. So I would have to divide 177147 by 4368 to get to all of the combinations for 1 inhabited planet per system for all of the combinations of 11 systems, which is the numerator, which is 40.6.

 

[blamocur]

So I have 16C11 for the number of possible ways I could chose 11 inhabited planets from 16 systems which is 4368,

It's systems, not planets there.
Once you picked a set of 11 systems you have to choose 1 out of 3 planets in each.

 

[Dr.Peterson]

So I would have to divide 177147 by 4368 to get to all of the combinations for 1 inhabited planet per system for all of the combinations of 11 systems, which is the numerator, which is 40.6.

No. Why do you think you would divide? Why do you think 40.6 is a reasonable answer?

What you should be trying to do at this point is to count the number of ways to select 11 planets, each from a different system. So you select 11 systems from the 16, and then select 1 from 3, 11 times. When you do one thing, and then do another, you use the multiplication principle, or fundamental counting principle.

 

[The Highlander]

Hi @Bluebell,

OK. All the correct “numbers” & “calculations” have already appeared throughout the thread but you still appear to be having a little difficulty in putting these together in the right “order”.

The time has now passed such that forum rules would allow me to post a complete solution to your problem but I would like to take one last opportunity to guide you to completing it yourself.

So, first, let’s just quickly review the “theory” required to solve your problem and stress some important points about it and how you need to use it.

The Maths is Fun website defines Probability (here) as…

​

Whenever I taught it, my preferred definition was…[math]Probability~of~any~Outcome(s)~happening =\frac{Number~of \textbf{\textit{Favourable}}~Outcome(s)}{Number~of \textbf{\textit{Possible}} ~Outcomes}[/math]
And I would rewrite that as just…[math]p=\frac{f}{n}[/math].

By “Favourable” I mean the Outcome (or Outcomes) that you are interested in happening (ie: just those you want to happen).

Now, it’s important to note (& understand why) \(\displaystyle n\) must always be greater than (or equal to) \(\displaystyle f\), so \(\displaystyle p\) is always a proper fraction, lying in the range 0 to 1, ie: 0 ≤ \(\displaystyle p\) ≤ 1.

Although \(\displaystyle p\) is always a proper fraction it is also (probably more often) expressed as a decimal fraction or a percentage fraction (and I think your teachers would probably expect you to put your final answer in the form of percentage in this case).

eg: the probability of tossing a head with a fair coin is ½ or 0.5 or 50%. Yes?

The next thing to note is that the complement of \(\displaystyle p\) is 1 – \(\displaystyle p\).

And while \(\displaystyle p\) is the probability of some outcome(s) happening the complement of \(\displaystyle p\) is the probability of it (or them) not happening (and it is often written as not \(\displaystyle p\)).

To go back to my tossing a fair coin example (see here), the probability of throwing a six is \(\displaystyle \frac{1}{6}\) so the probability of throwing any other number (1 – 5) is \(\displaystyle \frac{5}{6}\), ie: 1 - \(\displaystyle \frac{1}{6}\).

So once you work out the probability of something happening you can easily work out the probability of it not happening too (\(\displaystyle p\) and 1 – \(\displaystyle p\))

Now onto your particular problem…

The Question: "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?"

It may help your understanding and ability to reach the correct answer if you can visualize the situation, so I have made this diagram of the situation described in the problem…

​

This shows the 16 exosystems (which I have numbered 1 – 16) and three planets orbiting the Sun in each system. It also shows the 11 inhabitable planets (in green, below the exosystems) that will eventually replace some of the white planets currently shown in the exosystems. (The blue part of the Question.)

I am, henceforth, going to call the inhabitable planets “green planets” (because it’s easier to type) and the “white” planets I may refer to as “empty planets” or just “spaces” (because they represent a 'position' that a green planet might occupy).

Now I think we are all agreed that the best approach to this problem is to work out the probability that the eleven inhabitable planets are each in a different system, ie: there are exactly 11 systems containing a green planet. There cannot be more than 11 systems containing a green planet (because there are only 11 green planets) and there cannot be fewer than 11 systems containing a green planet (or at least one of them would then contain more than 1 green planet). Yes?

So the first problem is to work out the probability (\(\displaystyle p\)) that there are 11 systems each with one green planet in it, then the complement of that probability will give us the probability of that not being the case, ie: that there is at least one system with two or more green planets which is what the question actually asks you for (in the red part).

To do that we need to know the appropriate \(\displaystyle f\) and \(\displaystyle n\).

You quickly realized that you had to calculate the number of possible ways the green planets could be “fitted” into the 16 systems (this would give us our \(\displaystyle n\) for calculating \(\displaystyle p\)) and it was agreed by all that, since there were 48 (16×3) possible spaces they could go into, then this would be ₄₈C₁₁.

(NB: you expressed this as 2.26E10 but I would urge you not to write it down as that or enter into a calculator as that (because it is a rounded evaluation) instead just write it as ₄₈C₁₁ (or 48C11). When you actually calculate it just store it in memory as that will retain it as your best possible approximation to its actual value.)

So we now have…[math]p=\frac{f}{_{48}\text{C}_{11}}[/math]
And we just need to know the correct value for \(\displaystyle f\).

Again, you correctly figured out that, if we are only “interested” in 11 out of the 16 systems, the number of ways you could choose 11 systems from 16 was ₁₆C₁₁ and you worked that out to be (exactly) 4368 different ways.

Now I know I said previously that it didn’t matter how the green planets were arranged in the 11 systems they went into but that was pointed out as being incorrect and here is how I convinced myself that that was true.

Consider just one of those 4368 possible configurations. Let’s say, for instance, that one of the green planets is going into each of the first 11 systems, ie: those numbered 1 to 11 on the diagram.

In system 1 the green planet could go into the inner orbit or the outer orbit or the central one. But if we “fix” it in the inner orbit then there are 3 places it could go into in system 2, aren’t there? So that’s already 9 possibilities considering just the first 2 systems. Taking into account system 3 would multiply that by 3 again, now giving us 27 possibilities after considering just the first three systems! Therefore, in total, there are 177,147 possible arrangements of the green planets in just one of the 4368 systems we are interested in, which means there will also be the same number of possible arrangements in each of the 4368 systems!

So what will \(\displaystyle f\) be?

Can you now write out an expression for \(\displaystyle p\) in this form…

[math]p=\frac{_{16}\text{C}_{11}\times~?}{_{48}\text{C}_{11}}[/math]
Replace the “?” with the correct expression (?^?), work out \(\displaystyle p\) and then not \(\displaystyle p\) and then give your final answer as a percentage (to 2 or 3 decimal places)?

NB: You should state that your answer is to the number of places you have rounded it to, eg: "12.345% (to 3 d.p.)".

Looking forward to hearing from you soon (with the right answer). ?

 

[Bluebell]

Hi Highlander. Thank you for your really good explanation. I now understand where you are coming from. The order does make a difference so, the formula would be


p = (16C11 x 3^11)/48C11 = 4368 x 177147/2.259520036 E+10 (my calculator doesnt do combinations) = 0.0342452

and (1 - 0.0342452) = 96.575% (to 3dp)

The way I was thinking of it was like so

* * * * * * * * * * *
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48


The 16 triple planets are shown as 1- 48 arranged on top of each other in 3s to show each triple planet system. Right, so the position of the asterix's just show one way I could have selected 11 inhabited planets from 16 systems. The number of ways i could really arrange those asterix's along the top array of 16 triple planet systems is 4368. The position of each of those asterix's just represents one inhabited planet from the shown triple system, so isn't the numerator just 4368? It doesn't matter at all which of those 3 planets is inhabited. The asterix is equivalent to one inhabited planet?