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Percentages Question
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John Anderton
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RE: Percentages Question
quote:
Originally posted by Grue
Then don't post......
I was asking for clarification you nut :dodgy: Cant you understand ive been here long enough to know not to post if i didnt want to ...
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07-02-2006 09:21 AM
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Grue
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RE: RE: Percentages Question
[Pointless arguing]

quote:
Originally posted by John Anderton
I was asking for clarification you nut :dodgy:
You were not asking for clarification, you just stated the fact that you didnt get it. Reload2 clarified it perfectly for you already. This thread was not to help you understand but to help Roy with a problem.  No one would be happy if everyone on this forum posted "I don't understand" or "I don't get it" to every thread that is about a topic that they are less knowledgable about. If you think this okay can I go to every advanced programming thread and say "I don't get this."?

quote:
Originally posted by John Anderton
Cant you understand ive been here long enough to know not to post if i didnt want to ...

Its not about whether you want to post, but its about the forum rule and worldly annoyance-> SPAM. I have been here longer than you and I know that people don't like unhelpful spam.
[/Pointless Arguing]

Sorry about that, back on topic-> Problems of this type use the binomial model and some advanced calculators can do bernoulli trials such as this if you input the num of successes, num of trials, and the prob. of success. One calculator that I know can do this is the TI-83.
07-02-2006 09:43 AM
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.Roy
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O.P. RE: Percentages Question
Thanks a lot guys =D

Whats the formula for AT LEAST

For example whats the chance of getting AT LEAST 2 out of 5. When the chance of getting it once is 10%
07-02-2006 12:31 PM
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markee
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RE: Percentages Question
quote:
Originally posted by .Roy
Thanks a lot guys =D

Whats the formula for AT LEAST

For example whats the chance of getting AT LEAST 2 out of 5. When the chance of getting it once is 10%

Easiest way of doing this is working it out for 1 out of 5 trail with a chance of 10% and then taking this away from 100% as this is total probability.

So there are 5 possibilities (for the 1 out of 5 trails) ABBBB BABBB BBABB BBBAB BBBBA

1/10*9/10*9/10*9/10*9/10*5 = 32.805% of it NOT being at least 2 out of 5

100% - 32.805% = 67.195%

Chance of getting at east 2 out of 5 with a chance of getting it once being 10% is 67.195%. (This is so high because it is getting 2 out of 5, 3 out of 5, 4 out of 5 and 5 out of 5)




EDIT: Btw I checked this using my TI83+ :)

And @ Reload2

quote:
Originally posted by Reload2
quote:
Originally posted by Grue
that probability is 1/10*1/10*1/10*9/10*9/10=81/100000 so 81/100000*10= 81/10000 or .81% chance
I think that's the same formula that I wrote :P except I forget to convert to percentage... :P

You missed the *10 which makes a big difference, this is because there are 10 different possibilities you can make depending on what order you put it being successful and unsuccessful

This post was edited on 07-02-2006 at 12:51 PM by markee.
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07-02-2006 12:45 PM
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RaceProUK
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RE: Percentages Question
quote:
Originally posted by Grue
You were not asking for clarification
Do you not understand subtlety? Obviously not, so I'll be blunt.

When someone posts saying they don't understand, they want a sodding explanation, don't they?
quote:
Originally posted by John Anderton
I still aint clear about the question :dodgy:
This clearly implies he's looking for further details, so don't have a go at him, help him out!
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07-02-2006 02:15 PM
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bio_hazard13
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RE: Percentages Question
i cant really remember statistics from school but i realy seem to remember that with this thing, when you didnt have to worry about the order, you would times it by

{n
  x}

where n was the total number of trials (5 in this case) and x was the number of things that you were measuring (2 in this case). And i think the formula was

Probability of x times: {n x} x (probability of event happening)^x  x (probability of event not happening)^(n-x)

This post was edited on 07-02-2006 at 03:27 PM by bio_hazard13.
07-02-2006 03:23 PM
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.Roy
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O.P. RE: Percentages Question
quote:
Originally posted by markee
quote:
Originally posted by .Roy
Thanks a lot guys =D

Whats the formula for AT LEAST

For example whats the chance of getting AT LEAST 2 out of 5. When the chance of getting it once is 10%

Easiest way of doing this is working it out for 1 out of 5 trail with a chance of 10% and then taking this away from 100% as this is total probability.

So there are 5 possibilities (for the 1 out of 5 trails) ABBBB BABBB BBABB BBBAB BBBBA

1/10*9/10*9/10*9/10*9/10*5 = 32.805% of it NOT being at least 2 out of 5

100% - 32.805% = 67.195%

Chance of getting at east 2 out of 5 with a chance of getting it once being 10% is 67.195%. (This is so high because it is getting 2 out of 5, 3 out of 5, 4 out of 5 and 5 out of 5)




EDIT: Btw I checked this using my TI83+ :)

And @ Reload2

quote:
Originally posted by Reload2
quote:
Originally posted by Grue
that probability is 1/10*1/10*1/10*9/10*9/10=81/100000 so 81/100000*10= 81/10000 or .81% chance
I think that's the same formula that I wrote :P except I forget to convert to percentage... :P

You missed the *10 which makes a big difference, this is because there are 10 different possibilities you can make depending on what order you put it being successful and unsuccessful

Hey well, I understood what you said, but 67% doesnt make sense, because each one is 10% and its kind of weird to think that u have a 67% to get 2 or more, when getting one is 10%

I think the answer should be less then 10%
07-02-2006 04:17 PM
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Tochjo
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RE: Percentages Question
quote:
Originally posted by markee
Easiest way of doing this is working it out for 1 out of 5 trail with a chance of 10% and then taking this away from 100% as this is total probability.
Aren't you forgettting about the case 0 times out of 5?
07-02-2006 04:49 PM
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haydos
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RE: Percentages Question
The probabilty of getting at least 2 should be Probability of getting two (P2) + P3 + P4 + P5 (As said above).

A simpler way to determine it would be to work out the probability of one being picked plus the probability of none being picked and subtract this amount from one. I think markee's calculation was off after he didn't include the possibility of none being picked.
quote:
Originally posted by WDZ
don't be lazy
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07-02-2006 04:50 PM
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Grue
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RE: Percentages Question
quote:
Originally posted by haydn
The probabilty of getting at least 2 should be Probability of getting two (P2) + P3 + P4 + P5 (As said above).

A simpler way to determine it would be to work out the probability of one being picked plus the probability of none being picked and subtract this amount from one. I think markee's calculation was off after he didn't include the possibility of none being picked.

That is exactly right so working it out....
quote:
Originally posted by markee

So there are 5 possibilities (for the 1 out of 5 trails) ABBBB BABBB BBABB BBBAB BBBBA

1/10*9/10*9/10*9/10*9/10*5 = 32.805% of it NOT being at least 2 out of 5


And the probability of of it being 0 out of 5 would be 9/10*9/10*9/10*9/10*9/10 = .59049 or 59.049%
So 100%-59.049%-32.805%=8.146% which is less than 10% :)
07-02-2006 06:44 PM
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