Percentages Question - Printable Version
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Percentages Question by .Roy on 07-01-2006 at 06:28 PM
Whats The chance of you getting something 3 times out of 5
when the chance of getting it each time is 10%
Can you please give me the formula too?
RE: Percentages Question by Reload2 on 07-01-2006 at 06:39 PM
It should be: 1/10*1/10*1/10*9/10*9/10=81/100000
RE: Percentages Question by John Anderton on 07-01-2006 at 06:42 PM
quote:Overall chance of event = 3/5
Originally posted by .Roy
Whats The chance of you getting something 3 times out of 5
quote:Overall chance of event = 1/10
Originally posted by .Roy
when the chance of getting it each time is 10%
Why do i get the feeling i misinterpreted the question?
RE: Percentages Question by Reload2 on 07-01-2006 at 07:15 PM
Yes you misinterpreted it.
The question in other terms: What is the chance to get it three times on fiveif the chance to get it once = 1/10 ?
So the three times you get it * The two times you won't get it: (1/10*1/10*1/10)*(9/10*9/10)
RE: Percentages Question by absorbation on 07-01-2006 at 07:21 PM
0.6*0.1? If I am not right I have failed my statistics GCSE last Thursday 
.
RE: Percentages Question by RaceProUK on 07-01-2006 at 11:59 PM
Reload2 has posted the correct answer
RE: Percentages Question by .Roy on 07-02-2006 at 06:19 AM
quote:
Originally posted by Reload2
Yes you misinterpreted it.
The question in other terms: What is the chance to get it three times on fiveif the chance to get it once = 1/10 ?
So the three times you get it * The two times you won't get it: (1/10*1/10*1/10)*(9/10*9/10)
But it doesnt make sence, because if you multiply it by (9/10*9/10)
It will lower the chance, and I already know that if you get it 3 times in a row Its (1/10*1/10*1/10) Which SHOULD be a lower percentage then getting it in 3 random points of the draw.
RE: Percentages Question by Chrono on 07-02-2006 at 08:30 AM
quote:There's a slight difference though, if i understood your post correctly
Originally posted by .Roy
It will lower the chance, and I already know that if you get it 3 times in a row Its (1/10*1/10*1/10) Which SHOULD be a lower percentage then getting it in 3 random points of the draw.
. The probability of getting it 3 times in a row is indeed (1/10)^3. BUT, that's considering that you DON'T CARE about what you get in the other 2 tries. So, the probability is, being "A" what you want to get: P(getting A)^3 * P(getting whatever)^2 = (1/10)^3 * 1^2 = (1/10)^3 (Yes, the probability of getting it 3 times in a row or getting it AT LEAST 3 times in random points is exactly the same for that matter
).Now in the other question they ask you to calculate the probability of getting A 3 times OUT OF 5 (meaning that you have to get EXACTLY 3 times "A", not AT LEAST 3 times. So in the other 2 tries you MUST get, let's say, "B" which will obviously lower the possibility. So the probability is: P(getting A)^3*P(getting B)^2 = (1/10)^3*(9/10)^2
RE: Percentages Question by John Anderton on 07-02-2006 at 08:36 AM
I still aint clear about the question
RE: Percentages Question by Grue on 07-02-2006 at 09:09 AM
quote:Then don't post......
Originally posted by John Anderton
I still aint clear about the question
quote:I think you failed
Originally posted by absorbation
0.6*0.1? If I am not right I have failed my statistics GCSE last Thursday .
quote:
Originally posted by Reload2
It should be: 1/10*1/10*1/10*9/10*9/10=81/100000
No that is not quite right.... Lets call a success A(1/10) and a failure B(9/10). This equation above would be the probability of AAABB and not the probability of having 3 successes out of 5. You must consider all of the possibilities- AAABB AABAB AABBA ABAAB ABABA ABBAA BAAAB BAABA BABAA BBAAA
There are 10 possible different ways which you can have 3 successes and 2 failures. An easier way to find this number is just by multiplying 5(number of trials) by 2(number of failure/successes, whichever is the smaller number). Each of the 10 possible outcomes above all have the same probability of occuring and that probability is 1/10*1/10*1/10*9/10*9/10=81/100000 so 81/100000*10= 81/10000 or .81% chance
The formula would be...
(probability of successes^number of successes)*(probablility of failure^number of failures)*number of trials*(number of failure/successes, whichever is the smaller number)
* Grue hopes he is remembering his statistics right
RE: Percentages Question by John Anderton on 07-02-2006 at 09:21 AM
quote:I was asking for clarification you nut
Originally posted by Grue
Then don't post......
Cant you understand ive been here long enough to know not to post if i didnt want to ...
RE: RE: Percentages Question by Grue on 07-02-2006 at 09:43 AM
[Pointless arguing]
quote: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."?
Originally posted by John Anderton
I was asking for clarification you nut
quote: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.
Originally posted by John Anderton
Cant you understand ive been here long enough to know not to post if i didnt want to ...
[/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.
RE: Percentages Question by .Roy on 07-02-2006 at 12:31 PM
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%
RE: Percentages Question by markee on 07-02-2006 at 12:45 PM
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:I think that's the same formula that I wrote
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
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
RE: Percentages Question by RaceProUK on 07-02-2006 at 02:15 PM
quote:Do you not understand subtlety? Obviously not, so I'll be blunt.
Originally posted by Grue
You were not asking for clarification
When someone posts saying they don't understand, they want a sodding explanation, don't they?
quote:This clearly implies he's looking for further details, so don't have a go at him, help him out!
Originally posted by John Anderton
I still aint clear about the question
RE: Percentages Question by bio_hazard13 on 07-02-2006 at 03:23 PM
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)
RE: Percentages Question by .Roy on 07-02-2006 at 04:17 PM
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:I think that's the same formula that I wrote
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
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%
RE: Percentages Question by Tochjo on 07-02-2006 at 04:49 PM
quote:Aren't you forgettting about the case 0 times out of 5?
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.
RE: Percentages Question by haydos on 07-02-2006 at 04:50 PM
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.
RE: Percentages Question by Grue on 07-02-2006 at 06:44 PM
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%
RE: Percentages Question by .Roy on 07-02-2006 at 06:55 PM
Yeah Thanks So much, I thought something was wrong.
Thanks All
RE: Percentages Question by markee on 07-03-2006 at 05:44 AM
quote:
Originally posted by Grue
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%
Thanks for fixing that, I'm used to using nPr and nCr for things so I forget a bit of the binomial stuff, sorry about that .Roy.