math question

[busmaster11]

OK. One hundred pingpong balls in a lottery drum numbered one to a hundred. In a hundred turns, what is the probability you'll get the number "50" (just to pick an arbitrary number) exactly once?

How about atleast once?

How about twice? How about three times?

How about not at all?

And how did you calculate it?

 

[aphex]

Everytime the probability is 1/100 isnt it?

 

[busmaster11]

I think its 1/100 everytime, but that doesn't tell me the probability of getting a given number in 100 rolls. Yo0u can't just add them up to be 100% (1/100 * 100 turns = 100%) because you're still not guaranteed anything, though I would think its more than 50% likely...

 

[aphex]

But if your not removing any balls, the probability should stay the same i believe, no matter how many times you do it. Forgive me if im mistaken though, its been a few years since this type of stats

 

[aphex]

The more i think about it, the more i think im wrong

EDIT: Maybe not...

 

[911paramedic]

one out of a hundred, every time.

Just like flipping a coin, it is a 50/50 chance, each and every time.
If you flip it one hundred times and it comes up heads every single time, the odds it will come up heads again is still 50/50.

 

[aphex]

<< one out of a hundred, every time.

Just like flipping a coin, it is a 50/50 chance, each and every time.
If you flip it one hundred times and it comes up heads every single time, the odds it will come up heads again is still 50/50. >>



Thats exactally what i was thinkin... Maybe i was right after all

 

[busmaster11]

<< But if your not removing any balls, the probability should stay the same i believe, no matter how many times you do it. Forgive me if im mistaken though, its been a few years since this type of stats >>



It does stay the same for any given roll. Bujt somehow it can't add up to 100% if you do 100 rolls, because if you think about it, pick a number, and roll it 100 times. You're still not guaranteed to get the number you picked, even once.

 

[Theslowone]

it depends though, if you keep touching the ball then it will add fingerprints, which in turn will make the weight of the ball more and reduce the likely hood of it coming up again.

Not that it should be added into the problem. Just figured i would add confusion to the equation.

 

[busmaster11]

<< one out of a hundred, every time.

Just like flipping a coin, it is a 50/50 chance, each and every time.
If you flip it one hundred times and it comes up heads every single time, the odds it will come up heads again is still 50/50. >>



Okay, maybe I didn't make my question clear...

Let's start with the main question... In 100 rolls of the drum, how likely are you to get a given pre-selected number atleast once?

I know it is 1 out of 100 every time, but you have 100 shots at it. So whats the probability of getting it atleast once?

 

[Shalmanese]

Assuming without replacement, the chances of NOT getting a 50 is (99/100)^100 so it is 1 - that. to get 1 50 is 1/100*(99/100)^99 * 100

too tired to explain method

 

[busmaster11]

Okay... for those of you who still don't understand the question, how about this:

You are more likely to get "heads" at least once with two tosses of the coin compared to only one, right? So what are the chances of you getting "heads" atleast once with two tosses?

 

[911paramedic]

<<I know it is 1 out of 100 every time, but you have 100 shots at it. So whats the probability of getting it atleast once?>>

I am not a huge statistics person, but my answer remains the same.

If you roll your drum and pick a number and do not get yours, the next roll the odds are still 1/100, and so on. Each draw the odds are still 1/100. If you roll the drum, and do not get your number after 1,000,000,000,000,000,000,000,000 spins, the odds are still 1/100 your number will be drawn.

 

[busmaster11]

<< <<I know it is 1 out of 100 every time, but you have 100 shots at it. So whats the probability of getting it atleast once?>>

I am not a huge statistics person, but my answer remains the same.

If you roll your drum and pick a number and do not get yours, the next roll the odds are still 1/100, and so on. Each draw the odds are still 1/100. If you roll the drum, and do not get your number after 1,000,000,000,000,000,000,000,000 spins, the odds are still 1/100 your number will be drawn. >>



Look at my coin analogy. You're saying that infinite tosses of the coin won't increase your chances of getting "heads" atleast once, as compared to tossing it only once.

 

[Texmaster]

<< OK. One hundred pingpong balls in a lottery drum numbered one to a hundred. In a hundred turns, what is the probability you'll get the number "50" (just to pick an arbitrary number) exactly once?

How about atleast once?

How about twice? How about three times?

How about not at all?

And how did you calculate it? >>



A math question? I say we shoot him

 

[dopcombo]

erm, its a simple binomial distribution
n = 100 tosses
p = 1/100
X = prob of getting ball "50"

P(X>=1) = 1 - P(X=0) = 1 - 100C0 (99/100)^100

and similarly
key is understanding that it is a binomial distribution where the prob of getting a occurance is p and u've got 100 tries.

 

[busmaster11]

<< erm, its a simple binomial distribution
n = 100 tosses
p = 1/100
X = prob of getting ball "50"

P(X>=1) = 1 - P(X=0) = 1 - 100C0 (99/100)^100

and similarly
key is understanding that it is a binomial distribution where the prob of getting a occurance is p and u've got 100 tries. >>



So whats your answer?

btw what is "C" in your equation?

 

[busmaster11]

<<

<< OK. One hundred pingpong balls in a lottery drum numbered one to a hundred. In a hundred turns, what is the probability you'll get the number "50" (just to pick an arbitrary number) exactly once?

How about atleast once?

How about twice? How about three times?

How about not at all?

And how did you calculate it? >>



A math question? I say we shoot him >>



I'm sorry. Bush sucks, conservatives suck, the whole foreign policy sucks, we all suck... Now, back to my question.

 

[dopcombo]

oh 100C0 is 100 choose 0...er..

i dun really know how to explain it. been so long since i did the fundamentals

but ur calculator is definitely capable of this.

 

[Cattlegod]

i had statistics last semester, the answer is not 1/100, it is not 1/100 *100 either.






<< Assuming without replacement, the chances of NOT getting a 50 is (99/100)^100 so it is 1 - that. to get 1 50 is 1/100*(99/100)^99 * 100

too tired to explain method >>



this is on the right track, i didn't calculate it or think it through ( as a number or 2 may be off ), but this is how it is done.

 

[busmaster11]

<< oh 100C0 is 100 choose 0...er..

i dun really know how to explain it. been so long since i did the fundamentals

but ur calculator is definitely capable of this. >>



I think so too... But I don't know what 100 choose 0 is on the calc.

I think the correct answer for atleast once is more than 50% but less than 100%...

So no one can offer a percentage answer?

 

[Arschloch]

Here's what I have:

Probability of getting it at least once = 63.4%

 

[busmaster11]

<< Here's what I have:

Probability of getting it at least once = 63.4% >>



Sounds like the first plausible answer so far... So how'dya come up with it?

 

[Arschloch]

P(x>=1) = 1 - P(x=0).

The binomial functional form is P(x) = (n!/(r!(n-r)!)) * p^r * (1-p)^(n-r)
where n is the number of events (100, since you are drawing 100 times), r is the number of possible successful numbers (in this case, zero, as we are testing for the probability of getting #50 zero times), and p is the probability of success on any one trial (1% or .01, since there is a 1/100 chance of success).

Plug and chug... (100!/(0!(100-0)!)) * (0.01)^0 * (1-0.01)^(100-0)
(100! / 100!) * 1 * (0.99)^100 = .366 = 36.6%.

In other words, the chances of getting ball #50 zero times in 100 draws is equal to 36.6%. Then, 1 minus that result tells you the chances of getting ball #50 AT LEAST once. So, 1 - .366 = .634 = 63.4%.

Got it?

 

[bizmark]

Arschloch hat richtig geantwortet.

translation:

Asshole has answered correctly.