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Probability to be a terrorist when you have no facebook profile [Serious thread]

IAm

New member
Dec 18, 2012
3
Hello all,
I know it might sound harsh. I hesitated to disguise my question using another context but I hope some can analyse the exercice despite the emotion.

I would understand also if you can't. If mods think i am out of line, please delete the exercice.

Lets consider the following 3 hypothesis:
  1. t=Probability to be a terrorist= 10^-7
  2. p=Probability to not have a fb profile=10^-2
  3. The last 3 known terrorists had no facebook account.

With T=P(t/p) - Probability of being t knowing p
can we restrict the range of possible values of T with an 90% confidence ?

For example, with a 100% confidence, we know T is included between 0 and 100.t.

I have already got some partial results but i don't want to influence the way you may solve the question. I am still struggling to conclude anyway.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hello all,
I know it might sound harsh. I hesitated to disguise my question using another context but I hope some can analyse the exercice despite the emotion.

I would understand also if you can't. If mods think i am out of line, please delete the exercice.

Lets consider the following 3 hypothesis:
  1. t=Probability to be a terrorist= 10^-7
  2. p=Probability to not have a fb profile=10^-2
  3. The last 3 known terrorists had no facebook account.

With T=P(t/p) - Probability of being t knowing p
can we restrict the range of possible values of T with an 90% confidence ?

For example, with a 100% confidence, we know T is included between 0 and 100.t.

I have already got some partial results but i don't want to influence the way you may solve the question. I am still struggling to conclude anyway.
This reduces to the question: Given that the last 3 known terrorists had no face book account what can we say about \(P(p|t)\). The answer is nothing as we do not know how the information was selected.

For instance if we drew a random sample of size 3 from all known terrorists active in the last year and all three did not have fb accounts, then we could say something as we know how the data was generated.

CB
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Oh, geez, not here as well. What is it with this media-sponsored crusade against introverts and people who value their privacy? Correlation doesn't imply causation. I'm not offended though, it's just irritating that people even make this connection. At least you're trying to debunk this belief :)

I agree with CB on the math though, without knowing or taking a guess at the underlying distributions there's not much we can say.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Oh, geez, not here as well. What is it with this media-sponsored crusade against introverts and people who value their privacy? Correlation doesn't imply causation. I'm not offended though, it's just irritating that people even make this connection. At least you're trying to debunk this belief :)

I agree with CB on the math though, without knowing or taking a guess at the underlying distributions there's not much we can say.
I am sure I qualify as an quiet guy with mild Asbergers etc, but I do have a facebook account.
That means I am safe to be let out without supervision then...

What worries me is that the "norms" are trying to classify the nerdy as sociopaths and start a campaign to treat us, sorry you. I suspect that they are afraid of us, sorry again I mean you, for other reasons as well. Why does this remind me of X-Men?

CB Stands Up
CB: I am Magneto ...

(The solution that the US is looking for is called "Gun Control", and treating one another with respect might help as well)


CB
 
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IAm

New member
Dec 18, 2012
3
t is the probability to be a terrorist
p probability to have no fb profile

CB i am trying to find P(t/p): Probability to be a terrorist knowing he has no facebook.
(Not P(p/t): Probability to not have a fb page knowing he is a terrorist)


We know already that P(t/p) is somewhere between 0 and 100t
( P(t/p) is equals to 100t if All terrorists have no facebook page.)

I was able to calculate the probability that the last 3 terrorists have no fb depending on every value of P(t/p).
If P(t/p)=100t => Probability that the last 3 terrorists have no facebook is 100%
If P(t/p)=58t => Probability that the last 3 terrorists have no facebook is 20.54%
If P(t/p)=21t => Probability that the last 3 terrorists have no facebook is 0.93%

Indeed, the probability last 3 terrorists have no facebook is (P(t/p)/t*p)^3

Then i am a bit stucked.

(I am ok to add the hypothesis that the 3 terrorists we have found have been picked randomly if it is required to get a conclusion)
 

CaptainBlack

Well-known member
Jan 26, 2012
890
t is the probability to be a terrorist
p probability to have no fb profile

CB i am trying to find P(t/p): Probability to be a terrorist knowing he has no facebook.
(Not P(p/t): Probability to not have a fb page knowing he is a terrorist)
Bayes Theorem:- you need to know one to know the other:

\[P(t|p)=\frac{P(p|t)P(t)}{P(p)}\]

You know \(P(t)\) and \(P(p)\) where \(P(p)\) is the probability of not having a fb profile, \(P(t)\) is the probability of being a terrorist, both of which are givens.

\(P(t|p)\) denotes the probability of being a terrorist given no fb profile and \(P(p|t)\) is the probability of not having a face book profile given you are a terrorist.

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
t is the probability to be a terrorist
p probability to have no fb profile

CB i am trying to find P(t/p): Probability to be a terrorist knowing he has no facebook.
(Not P(p/t): Probability to not have a fb page knowing he is a terrorist)


We know already that P(t/p) is somewhere between 0 and 100t
( P(t/p) is equals to 100t if All terrorists have no facebook page.)

I was able to calculate the probability that the last 3 terrorists have no fb depending on every value of P(t/p).
If P(t/p)=100t => Probability that the last 3 terrorists have no facebook is 100%
If P(t/p)=58t => Probability that the last 3 terrorists have no facebook is 20.54%
If P(t/p)=21t => Probability that the last 3 terrorists have no facebook is 0.93%

Indeed, the probability last 3 terrorists have no facebook is (P(t/p)/t*p)^3

Then i am a bit stucked.

(I am ok to add the hypothesis that the 3 terrorists we have found have been picked randomly if it is required to get a conclusion)

This is invalid in that your last 3 so called terrorists is not a random sample of size 3 of all active terrorists.

Also your probability of not having a fb profile is ludicrous, it is not the case that 99% of people have fb profiles.

In fact if we take crude statistics the probability of a random human not having a fb profile is ~=86%, and probably around 40% if you really want to talk about US residents. Also real terrorists are concentrated in parts of the world where the proportion of people without fb accounts is higher.

You might also note if we are really talking about spreed killings perpertrators rather than "terrorists" then Ander Beivik did have a fb profile.

CB
 
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IAm

New member
Dec 18, 2012
3
This is invalid in that your last 3 so called terrorists is not a random sample of size 3 of all active terrorists.
I thought this was valid as long as having a fb account does not impact the capability of identifying a terrorist.

In other terms,
Probability of identifying a terrorist = "Probability of identifying a terrorist with a fb account" ="Probability of identifying a terrorist without a fb account"

Regarding the other hypothesis (% of fb account and the term "Terrorist") you are right but it was just to get an actual figures for the analysis.

Thank you CB for your help.