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Topic: 5 counters in a bag (Read 911 times) |
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Noke Lieu
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5 counters in a bag
« on: Nov 3rd, 2014, 2:37am » |
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Aw heck, make them marbles if you like...
At least 3 of them are black, any others are white.
What's the probability that the 3rd counter you draw from the bag (without replacement) is black?
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Grimbal
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Re: 5 counters in a bag
« Reply #1 on: Nov 3rd, 2014, 3:28am » |
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"At least 3", does it mean the number of black counters is uniformly distributed between 3, 4 and 5?
Either the quesiton is very simple or I am missing something.
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towr
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Re: 5 counters in a bag
« Reply #2 on: Nov 3rd, 2014, 9:57am » |
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at least 60%
In any case, it doesn't matter whether you look at the third or the first or the Kth.
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Noke Lieu
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Re: 5 counters in a bag
« Reply #3 on: Nov 3rd, 2014, 6:10pm » |
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on Nov 3rd, 2014, 3:28am, Grimbal wrote:| Either the quesiton is very simple or I am missing something. |
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Just a simple question (it is in easy)... possibly the trap was trying to get you to overthink it... 
Yes I was intending there to be either uniformly 3, 4 or 5 black counters.
There's one part of me that wonders about conditional probability- only because I fear that it crops up more than it actually does.
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Noke Lieu
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Re: 5 counters in a bag
« Reply #4 on: Nov 9th, 2014, 8:59pm » |
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ah well...
It was something that I might have messed up, bt I was interested in a quirk...
When I was a younge man, my teacher told me that if in trouble or in doubt, write the sample space out. Not terribly practical in quite a few situations, but not the worst adivce either...
bbbww[/b]
bbwwb
bwwbb
wwbbb
bbwbw
bwbbw
wbbbw
bwbwb
wbbwb
wbwbb
bbbbw
bbbwb
bbwbb
bwbbb
wbbbb
bbbbb
which is 11/16 successes.
But 1/3(1+6/10+4/5) = 12/15
and 11/16 =/= 12/15...
so what's going wrong?
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| « Last Edit: Nov 9th, 2014, 9:01pm by Noke Lieu » |
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towr
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Re: 5 counters in a bag
« Reply #5 on: Nov 9th, 2014, 10:37pm » |
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on Nov 9th, 2014, 8:59pm, Noke Lieu wrote:The samples aren't equi-probable. It's like saying "what's the chance of throwing a 1 with a die" and describing the sample-space as:
* one
* not-one
and coming up with an answer of 1/2.
At the very least, the three sample-subspaces should be weighed equally (if 0, 1 or 2 white tokens are in fact equally likely), and not weighed by 1, 5 and 10.
The easiest way to fix it so you get the right sample-space, is to give the two extra tokens an identity. So e.g. use permutations of bbb12
Actually, that might still suggests a different sample-space than you might expect. If each of the two extra token has a 50% chance of being black, the cases of 0, 1 or 2 extra black tokens are not equi-probable; having 1 extra black is twice as likely. (But ultimately the answer it gives is the same)
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Noke Lieu
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Re: 5 counters in a bag
« Reply #6 on: Nov 10th, 2014, 1:48am » |
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y'know, I know that, but can't see it clearly enough to trust myself... One day, one day... 
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