As the original questioner, I hesitate to answer this, but . . . .
a) At the start, there were 700 green, 1300 red, not the other way around
b) if the probability that at least half the beads chosen are green is vanishly small, then the probability that at least half the beads are red is indeed almost one.
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>if the probability that at least half the beads chosen are green is vanishly small, then the probability that at least half the beads are red is indeed almost one
This is the part that does not make sense to me. The probability of having 250 green ones cannot be vanishly small There are 700 green and 1300 red. You select 500 at random. If the probablity of selecting at leat 250 red is 1 then this means that you WILL select 250 red ones. What about the other 250? They must be green. Intuitively this does not seem right. I cannot work out how Vin worked out the result. It would be interesting to pose the question the other way around and calculate the probability of 250 reds using the same method
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If the probablity of selecting at leat 250 red is 1 then this means that you WILL select 250 red ones. What about the other 250? They must be green.
This is what's confusing you. You're forgetting the words AT LEAST. Yes, it's almost certain that 250 will be red, but this doesn't mean that the other 250 WILL be green, it's AT LEAST 250 red, so many of the other balls can also be red.
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The chance of being killed by lightning in the United States is equal to about 0.00000032
Even less if you never go there! ;-)
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L\'escargot.
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Robin,
This is calculated by working out the probability of having exactly 250 green ones, adding the probability of having 251 green ones, etc, etc. These add up to the number above.
The number that adds up to very nearly one is the odds that there are somewhere between 1 and 249 green ones in the sample of 500. What the results say is that this is almost a certainty. Thus, it's saying that the odds of >249 (i.e. 250 red ones or 251 red ones or 252 red ones, etc etc) is 1 as near as dammit.
Don't know if that clarifies matters or murks them.
V
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Put another way. If the probability of >=250 green is X, then the probability of =250 is around zero, so probability <250 is around 1.
V
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I have 50 jars, each with 40 beads - 14 green, and 26 red. From each jar, I take (at random) 10 beads, and put them in a bowl, which thus has 500 beads. What is the probability that the number of green beads outnumbers the red ones?
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tyro : can you tell me whether
1.
you take 10 beads out of each jar in a single event, and then repeat this for the 50 jars, or
2.
take 1st bead out of the 40 in the first jar, repeat this for the other 49 jars;
take 2nd bead out of the remaining 39 beads in that jar, repeat this for the other 49jars;
take 3rd bead of the remaining 38 beads in the first jar, repeat this for the other 49 jars; etc. etc. until
take 10th bead of the remaining 31 beads in the first jar, repeat this for the other 49 jars.
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To get to the bottom of this one I think we need someone who knows how many beans (or beads) make five!
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L\'escargot.
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To get to the bottom of this one I think we need someone who knows how many beans (or beads) make five! -- L\'escargot.
I know the answwer to that:
a bean, a bean, a half a been, a half a bean, a bean, a bean.
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Dalgish - why do you think that will make a difference?
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tyro : can you tell me whether 1. you take 10 beads out of each jar in a single event, and then repeat this for the 50 jars, or
Yes, 1. Why - does it make a difference?
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I think that this problem is alot more complicated than it first appears.
Trying to model the answer I end up with some huge calculations, but it is very clear that the probability of achieving the answer by chance is as close to Nil as any normal person would want to get to! Orders of magnitude smaller than the earlier answers.
However you cannot tell from the question if the chances of picking each individual bean from every is really only dependant upon the occurence in the closed population. For example are green beans found closer to the top as red beans are smaller etc This would have a significant impact upon the answer.
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As I'm utterly hopeless with stats, I adopted a brute force approach, and simulated the problem directly. The results of 1139717 trials are shown below (I intended to do more, but the simulation was taking longer than I have time for, so I interrupted it, hence the peculiar number of trials)
Below is a histogram of results, so, for the first non-zero case, there were 6 trials which had green bead counts between 130 and 135, 43 trials giving between 135 and 140 beads, and so on.
120 0
125 0
130 6
135 43
140 415
145 2740
150 11896
155 39086
160 94602
165 169925
170 229570
175 234590
180 180556
185 106405
190 47577
195 16771
200 4472
205 922
210 120
215 20
220 1
225 0
230 0
235 0
240 0
245 0
250 0
As previously stated, the probability of getting more than 250 beads in a trial of 50 jars is very small indeed. Although the probability of getting equal to or more than five beads from any particular jar is much higher, approximately one jar in five.
Number_Cruncher
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Being a sad ******* i have sat down and worked this out:
The results are as follows:
1 Jar 22%
2 Jars 9%
3 Jars 4%
50 jars will take a little longer!
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The maths is:
If your probability of picking a green bean is P, then the probability of picking exactly m green beans in a selection of exactly n is:
nCm x P^m x ( (1-p)^(n-m) )
nCm = n!/(m!x(n-m!))
n! = factorial of n
^ = to the power of
Plug in n=175, m=500 p = 0.35 and you'll find that the probability is 0.0.03738 - so nearly 4% of the time you'll pick out 175 green beans exactly.
All maths courtesy of Collin Phillips, University of Sydney.
V
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Obvious error:
nCm = n!/(m!x(n-m!)) should read nCm = n!/(m!x(n-m)!), obviously.
V
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But unfortunately that was not the question! It was:-
You have 50 jars each containing 40 beans, 16 green and 24 red
You pick 10 beans from each jar and put them in a bowl, totalling 500 beans
What is the probability that you end up with 250 or more green beans in the bowl.
It is the combination of the beans from each jar that is causing me some problems. For example using 3 jars you could get 15 beans or more (the equivalent percentage) by having 0 beans from jar 1, 7 from 2 and 10 from jar 3. or 3,4,8 etc etc etc
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The maths posted by Vin does break the back of the problem, because it allows you to estimate the probability of obtaining 250, 251, 252, ..... 499, 500 green beads, and then, to obtain the final answer, one may sum these probabilities. If you include the case for 250 beads, the sum of probabilities comes out at 4.39E-12 (or, 1 case in 228049112213). Practically, it's fairly pointless adding terms beyond 260, because they are so improbable, they add little to the total probability.
Vin - please could you post a link to the web page?
Number_Cruncher
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It is the combination of the beans from each jar that is causing me some problems. For example using 3 jars you could get 15 beans or more (the equivalent percentage) by having 0 beans from jar 1, 7 from 2 and 10 from jar 3. or 3,4,8 etc etc etc
I'm a little confused by your description:
He said you take ten beans from each jar, surely? If each jar contains a proportion of 35% green, 65% red, then if you take them at random in any combination that doesn't exceed the number of one or another colour, then you'll end up with a suitable sample. (What I mean by that is that if a jar of 40 contains 14 green and 24 red, then if you take a sample from each of 25, you are guaranteed a green in your final sample)
The link to the relevant page is:
tinyurl.com/y2799s
V
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