Probability question

If 1 in 100 people wear glasses, then what is the probability that I, who is in a group of 5000 people, wears glasses?

With only that information, one in a hundred?

Where did you get your facts? I don't think they are reliable.

I wear a monocle and I'm in ur group, killin your stats

gawd, you guys suck.

42

5000/100 = 50

Your probability is 1/50 assuming uniform random disribution and that there are no other factors that you haven't mentioned.

>>7
is right. But wearing bad eyesight is genetic and it depends more on whether or not your folks and granparents have bad eyesight and wear glasses.

Does wearing contacts count?

OK, if you want a serious reply, I'd say the original question is simply meaningless as stated. "1 in 100 people wear glasses" means nothing if you don't specify the population it applies to, and the probability you are asking for entirely depends on the "group of 5000 people" you consider (if you pick 5000 near-sighted people among the general population, the probability will likely be 1, irrespective of what the ratio is for the general population).

this thread of made out of retard

>>7
Thanks, thant's what I wanted. I know nil about probability and I was a bit confused about that and I didn't want to pore through long explanations to get the answer at page 2345234.

Hay, this is the OP. The only person who got it right was >>2.

I made the facts up on the spot. I did this because I am a homosexual and your stupidity gives me a hard on. I am currently jerking off.

1 in 100. Logic fails thread

>>12,13
okay that makes more sense i guess.

>>7
>>8
You fucking morons. 1 in 100 wear glasses, so 50 in 5000 wear glasses. The answer is clearly 1/100 regardless of the group.

Think of it this way. You're in a room with 99 other people. What are the odds that you're wearing glasses? Now 4900 more guys enter the room. Have your odds suddenly changed?

Idiots. Sage.

>>12
Nope, the only right answer was >>9, because he knows about universes and conditional probabilities. The original question was indeed meaningless as stated.

>>9 is an idiot because the OP question obviously had nothing to do with the real world. It was a simple dumb question and the obvious assumptions should have been gleaned from it.

>>17
Truth

>>16
Saying something is "meaningless as stated" is just a bullshit way to get out of answering the question.

Jesus some people are morons.

OP: The probablity of Person X wearing glasses in a group of three billion trillian is 1/100. Simple as that.

Oh, and what about the probability of person X wearing glasses in a group of ten people? Let me chuckle if you still say 1/100. Belonging to the group cannot be independant from wearing glasses in this case, and need not be in the OP question.

YOU ARE ALL IDIOTS FOR CONSIDERRING QUESTIONS THAT HAVE NOTHING TO DO WITH THE REAL WORLD

EITHER DO HYPOTHETICAL MATH OR ACTUAL SCIENCE

HYPOTHETICAL SCIENCE BY DEFINITION IS NOT SCIENCE

>>1
http://dis.4chan.org/read/sci/1157719450/1-*

FUKKEN FIX'D RETARD

http://dis.4chan.org/read/sci/1157719450/1-

>>20

it is 1/100. (disregarding the other errors.)

Let me chuckle if you still say 1/100.
Holy fucking shit, you are retarded.

>>20
Yes, it's 1/100 you fool. gb2/statistics/ or kill yourself, kthxdie.

1/5000?

>>24, >>25, >>26
Uhuh. A uniform probability of 1/100 in a group of *ten* people? So, tell me, how many of those ten people actually do wear glasses? You don't seem to have any idea of what "probability" actually means, even though discrete probabilities are little more than glorified kindergarten arithmetic.

So, tell me, how many of those ten people actually do wear glasses?
Answering that question adds information to the problem, and only then does the probability change. However, with only the information that 1 in 100 people wear glasses, the answer to the question is not known. All ten people could wear glasses (although this is the most unlikely option: .01^10), or none could. You don't know!

You're supposed to learn this stuff in high school (so I guess you're probably trolling).

>>28
The expected number of people who wear glasses in a group of 10 is 0.1 (assuming we have no other information about the group of 10).  Consider a similar situation: How many numbers in the set {k, k+1, k+2} (k an integer) are divisible by two? Without knowing k, our answer is 1.5. The reason for this is that either 1 of the numbers must be even, or 2 must be even, and these possibilities occur with equal probability, so the expectance is (0.5)*1 + (0.5)*2. If, on the other hand, we know k is even (or odd) then {k, k+1, k+2} has 2 (1) even numbers.

Also, I'd like to say that it's kind of humorous that you should bring up kindergarten arithmetic, since your mathematics education does not seem to have gone past whole numbers.

>>30

lol 2 divides 1.5 of these numbers. you fucking fail. the counting branch of mathematics only works in whole numbers.

>>31
You don't seem to have any grasp of Expectance. Fail.

I'm not trolling. I'm just a bit surprised at how confused people are when it comes to probabilities.

You're really confusing a stochastic (in this case Bernoulli) *process* with the probability of a simple *event*. The probability of someone wearing glasses in a group of N people, n of whom do wear glasses is just p=n/N (and hence certainly cannot be 1/100 when N=10).

Now if you consider the experiment of picking a person at random (i.e. uniformly) from the lot and checking whether she wears glasses, it's a Bernoulli process of parameter p (which again cannot be 1/100 when N=10, that's something I *do* know). You can indeed compute the probability of getting k successes (i.e. people wearing glasses) after N iterations of that process, but that's pretty irrelevant to the whole problem.

>>33
We have a group of n*100 people for some integer n. We have p = 1/100 (or n/(n*100) if you prefer). Now we say that there's some group of 10 people, a subset of the n*100 people. What is the chance that one of the people in this group of 10 wears glasses? It's still 1/100 for each person in the group. The group may have 0 people who wear glasses, or 10 people who wear glasses, but each person in the group wears glasses with probability 1/100. If you leave out the part about the 10 people being a subset of the n*100 people then you are correct, the probability that a person in that 10 person group wears glasses cannot be 1/100. If you keep the 10 person group a subset of the n*100 person group and add more information (such as "8 people in this ten person group wear glasses") then the expectance that someone in the group wears glasses is not 1/100. But your answer to the question as stated (10 person group is a subset of n*100 person group and probability of someone in the n*100 person group wearing glasses is 1/100, what is the probability that a person in the 10 person group wears glasses?) is not correct.

>>33
is just p=n/N
No. This is how you'd calculate p if n were known, which it isn't. Also, you wouldn't even get the p we were given (which is the chance that someone wears glasses in general) but only an approximation of it, based on the composition of the group (which, again, could be anywhere from 0 to 10 people with glasses).

>>30
(This is >>28 and >>33 speaking).

It's kind of funny you should take this precise example, because, as you should know, there is no such thing as a (non-compactly supported) measure on the set of non-negative integers, and hence you cannot really talk about the expected value of the random variable you consider here. Also, nobody calls that an "expectance".

But you do have a point: it is true that if you pick k people uniformly in a group of N, n of whom where glasses, the expected number of people wearing glasses among those you've picked is nk/N. So the *mean* probability of one people wearing glasses among those you picked, when averaged over *all* possible picks, is n/N. But that certainly was not the question (an expected value is not the same thing as a probability). It says nothing about what happens in any given group, so that, given no additional information, we simply tell. Probably >>16's point.

Regarding my mathematical education, I can safely say it extends a bit further past whole numbers, although it revolves around them in some sense.

>>35
I don't know what to say. Lookup the word "probability" or something.

I don't know what to say.
Good.

>>36
"as you should know, there is no such thing as a (non-compactly supported) measure on the set of non-negative integers"

First off, I believe you forgot to include the adjective translation-invariant. There certainly is a non-compactly supported measure on the set of non-negative integers. Let mu({k}) = 2^(-k-1). Then mu({0,1,2,...}) = 1, making ({0,1,2,...},P({0,1,2,...}),mu) not only a measure space with a non-compactly supported measure, but also a probability space.

Regardless, your point that there isn't any uniform random distribution on the non-negative integers is well taken. I shouldn't have cut corners in that example, but it can easily be rectified: Let k be uniformly distributed in the integers modulo 2.

THIS THREAD IS FUCKING STUPID

THE PROBABILITY IS THE SAME, IT'S ALWAYS THE SAME PER PERSON, THAT'S WHAT PROBABILITY IS

>>39
I stand corrected, I did mean translation-invariant rather than just non-compactly supported.

Anyway, to clarify my point, let me suggest alternate formulations of the OP question:

1/ Consider a group of 5000 people, one in 100 of whom wears glasses. Pick one at random (uniformly): what's the probability of her wearing glasses?
2/ Consider a group of N>5000 people, one in 100 of whom wears glasses. Pick 5000 of them at random (uniformly): what's the expected probability of one those wearing glasses?
3/ Consider a group of N>5000 people, one in 100 of whom wears glasses. Pick 5000 of them: what's the probability of one those wearing glasses?
4/ I belong to a group of N>5000 people, one in 100 of whom wears glasses. Pick 5000 of them at random, myself included: what's the expected probability of one those wearing glasses?


1 and 2 have 1/100 as an obvious and definite answer. You can't say anything about 3. And the answer to 4 depends on whether "I" wear glasses or not. That's only a few possible variations.

If 1% of people wear glasses, there is a 1% chance that one person in a randomly selected group of 5000 wear glassss.

If 1% of people wear glasses, there is a 1% chance that a person in a randomly selected group of 5000 wear glasses. *

fix'd

there a a fuck lot of confused statistics guys here. let me guess: economics?

>>4 won