Probability & factorials

[PeddlerOfBombast]

Hi, folks... Quick question, with my old stats books in storage, thought I turn to y'all...

Let's say the total population is 1000. There are 100 left-handers in the population. If I were to randomly choose 50 people, what's the probability that all 50 would be Southpaws? I know it's relatively simple, so thanks for indulging me. Let me know! Thanks...

 

[Deleted member 4993]

Hi, folks... Quick question, with my old stats books in storage, thought I turn to y'all...

Let's say the total population is 1000. There are 100 left-handers in the population. If I were to randomly choose 50 people, what's the probability that all 50 would be Southpaws? I know it's relatively simple, so thanks for indulging me. Let me know! Thanks...

If you choose 1 person - what's the probability that that one would be Southpaw?

If you choose 2 persons - what's the probability that those two would be Southpaws?

If you choose 3 persons - what's the probability that those three would be Southpaws?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

 

[PeddlerOfBombast]

If you choose 1 person - what's the probability that that one would be Southpaw?

If you choose 2 persons - what's the probability that those two would be Southpaws?

If you choose 3 persons - what's the probability that those three would be Southpaws?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

All apologies, I was just looking for a quick answer... I'm 47, far outside of lessons, & writing an article in which I need a probability equation given the above example. I admittedly didn't read-before-posting... oops... so will take my question elsewhere. I now deduce that you're more geared towards helping students through mathematics rather than a scrappy bunch of fellow nerds who like to banter about numbers & answer silly questions. Sorry again, good for you all (I come from a long line of educators), and best of luck. Stay safe! Cheers.

 

[pka]

Let's say the total population is 1000. There are 100 left-handers in the population. If I were to randomly choose 50 people, what's the probability that all 50 would be Southpaws? I know it's relatively simple, so thanks for indulging me.

We need to understand that these answers are restricted to this one population.
There are $\dbinom{1000}{50}$ ways to choose fifty from one thousand. That is a gigantic number.
Of all those there are $\dbinom{100}{50}$ ways to choose fifty from all the left-handed from one hundred. That is also a gigantic number.
If we divide the two we get $1.0664527273885084246767172977907908171174510845558287 × 10^{-56}$.
That is as close to zero as I have ever seen.

 

[JeffM]

In case you are not used to the notation

[MATH]m = 0 \implies m! = 1;\\ m \text { is an integer} > 0 \implies m! = m * (m - 1)!\\ n \text { and } k \text { are integers such that } n \ge k \ge 0 \implies \dbinom{n}{k} \equiv \dfrac{n!}{k! * (n - k)!}.[/MATH]
[MATH]\therefore \dfrac{\dbinom{100}{50}}{\dbinom{1000}{50}} =[/MATH]
[MATH]\dfrac{\dfrac{100!}{50! * 50!}}{\dfrac{1000}{50! * 950!}} =[/MATH]
[MATH]\dfrac{100!}{\cancel{50!} * 50!} * \dfrac{\cancel{50!} * 950!}{1000!} =[/MATH]
[MATH] \dfrac{100! * 950!}{50! * 1000!} = \dfrac{100}{1000} * \dfrac{99}{999} \ .... \approx 0.[/MATH]

 

[pka]

In case you are not used to the notation

[MATH]m = 0 \implies m! = 1;\\ m \text { is an integer} > 0 \implies m! = m * (m - 1)!\\ n \text { and } k \text { are integers such that } n \ge k \ge 0 \implies \dbinom{n}{k} \equiv \dfrac{n!}{k! * (n - k)!}.[/MATH]
[MATH]\therefore \dfrac{\dbinom{1000}{50}}{\dbinom{100}{50}} =[/MATH]
[MATH]\dfrac{\dfrac{1000!}{50! * 950!}}{\dfrac{100}{50! * 50!}} =[/MATH]
[MATH]\dfrac{1000!}{\cancel{50!} * 950!} * \dfrac{\cancel{50!} * 50!}{100!} =[/MATH]
[MATH] \dfrac{50! * 1000! }{950! * 100!} \approx 0.[/MATH]

You have things up-side-down. It is $\dfrac{\dbinom{100}{50}}{\dbinom{1000}{50}}$

 

[JeffM]

You have things up-side-down. It is $\dfrac{\dbinom{100}{50}}{\dbinom{1000}{50}}$

Yep. I was editing while you were correcting.

 

[PeddlerOfBombast]

That's it! 1989 math class is slowly coming back. Thanks for the help, folks! And again, great work you do! Cheers...

 

[PeddlerOfBombast]

Hi again... Thanks for all your help so far.

Full disclosure, I'm trying to determine how statistically improbable it is for someone to randomly 'friend' a given number of profiles, out of all the profiles online (of a specific age range), that just happen to be 'decoy' profiles. I know it's a hugely small number, but that's my point. (This is a sensitive topic that you can probably discern, so that's all the detail I'll go into here; if there's a way of making this thread private or hidden, please feel free to do so.)

So, say there are 65M people in the UK, 20% of which (13M) are in Age Group (A). Out of those, say 1/3 (4M) have online profiles.

Out of those 4M profiles (the total set to consider), let's say 1,000 are decoys.

If a person randomly friended 3 profiles, what's the probability that all 3 would be decoys?

Based on your previous advice, that'd be 1000(choose)3, divided by 4M(choose)3 - correct? (Apologies for the crude notation.)

Now, here's the kicker. Say the 1000 decoy profiles are run by 125 'teams' with 8 decoys each.

What's the probability of 3 random profiles not only being decoys, but all from 1 specific team? It's gotta be statistically mind-boggling, but again, that's the point.

Please let me know if I'm thinking of this correctly, and if you can, what those probabilities are. (If there's an online calculator into which I can plug these values, please do let me know.)

Thanks so much again. This info will be used to strengthen real cases and real work being done to keep the online community safe. Your help so far is greatly appreciated! Cheers...

 

[JeffM]

The actual computation of your revised problem on the ASSUMPTION of random selection is

[MATH]\dbinom{8}{3} \div \dbinom{4,000,000}{3} = \dfrac{8!}{3! * 5!} \div \dfrac{4,000,000!}{3! * 3,999,997!} = [/MATH]
[MATH]\dfrac{8 * 7 * 6}{3!} \div \dfrac{4,000,000 * 3,999,999 * 3,999,998}{3!} =[/MATH]
[MATH]\dfrac{16 * 21}{4,000,000 * 3,999,999 * 3,999,998} < \dfrac{1}{10^{20}} \approx 0.[/MATH]
The problem is that the assumption of random selection makes absolutely no practical sense. Decoys are chosen to be anything but randomly appealing.

EDIT: Lures are directed at specific audiences, which are much smaller than your base. The probabilities will still be low, but much, much greater than what the computation above shows. My billing rate is $300 per hour. Shall I send you a letter of engagement?

 

[PeddlerOfBombast]

Dang, Jeff! I certainly owe you a pint after that.

To your point on the base assumption of randomness, decoys are specifically designed to be neutral - that is, not specifically targeted / appealing to a specific person/group, outside of age & location, the latter of which is covered under the total set. They're just listed as name, age, location, & picture, meant to be a wide generalization. True, given the circumstances (which I presume you've discerned), there may be certain factors of the decoys that appeal to (or target) certain people, but everything is done to make them just another face in the crowd, as it were.

Anyway, thanks again. Your help is immensely appreciated, & will help in some great things.

And, I should say - getting back to chatting maths is making me want to pick up my old textbooks. I love (re)learning a great many things, and this episode has triggered my mathlete muscles. Thanks for that, on a personal note. Cheers!

 

[PeddlerOfBombast]

(Also, any idea how to handle the teams issue? Your solution tackles the overall number of decoys, but not the probability of selecting 3 from the same team. That's beyond my recalled capabilities at this point. And are there any online calculators into which these figures can be plugged? It'd help for future uses. Do let me know if you can. Cheers!)

 

[JeffM]

(Also, any idea how to handle the teams issue? Your solution tackles the overall number of decoys, but not the probability of selecting 3 from the same team. That's beyond my recalled capabilities at this point. And are there any online calculators into which these figures can be plugged? It'd help for future uses. Do let me know if you can. Cheers!)

I was afraid you might ask that.

Actually my computation was for picking three from a single team. Someone like pka or romsek who are whizzes at this branch of math can probably give you an exact formula easily, a formula that I suspect will be a bit complex, but the formula will probably result in a number that is, for practical purposes, virtually zero.

The probability that, selecting at random, you will pick three from any combination of teams is

[MATH]\dfrac{(1000)!}{3! * 997!} \div \dfrac{(4 * 10^7)!}{3! * (4 * 10^7 - 3)!} =\\ \dfrac{1000 * 999 * 998}{3 * 2} \div \dfrac{4,000,000 * 3,999,999 * 3,999,998}{3 * 2} \approx\\ \dfrac{10^9}{64 * 10^{21}} < \dfrac{1}{10^{13}}.[/MATH]
And the probability that you will pick all three from the same team given that you have picked three from the thousand available is

[MATH]\dfrac{7}{999} * \dfrac{6}{998} = \dfrac{2}{333 * 244} < \dfrac{2}{100 * 1000} = \dfrac{2}{10^6}.[/MATH]
So the probability that you will pick three from a single one of the 125 teams is less than

[MATH]\dfrac{1}{10^{13}} * \dfrac{2}{10^{6}} = \dfrac{2}{10^{19}}.[/MATH]
Understanding that is an upper bound, I'd say the probability is remote. I'd bet a lot against it happening in my lifetime (death cancels all obligations according Zuleika Dobson), and professionally I was a banker.

 

[PeddlerOfBombast]

Thanks again! And if anybody does have more info / formulae / calculators available, please do let me know.

The remoteness of possibility, again, is the point. We have individuals saying that it's complete coincidence that they chose X (eg. 3) random (but in the subset) profiles that all happen to be in the same team. So, either that's true - and why I need the probability - or, they've not JUST targeted the 3 that we know about, but in fact many more. It's an interrogation point, & I like specificity. (I want to compare it to DNA probabilities.) Cheers again, friend!