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 | | From: | grillatzo at hotmail.com | | Subject: | PROBABILITIES | | Date: | 23 Jan 2005 15:06:03 -0800 |
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 | Suppose you are truthfully told that ten marbles were inserted into a
box, all of them identical except that their colors were determined by
the toss of an unbiased coin. When heads came up, a white marble was
inserted, and when tails came up, a black one. You reach into the box,
draw out a marble, inspect its color, then return it to the box. You
shake the box to mix the marbles randomly, and then reach in and again
select a marble at random. If you inspect ten marbles in succession in
this manner and all turn out to be white, what is the probability to
the nearest whole percent that all ten marbles in the box are white?
Regards
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| | From: | David Kastrup | | Subject: | Re: PROBABILITIES | | Date: | Mon, 24 Jan 2005 01:17:37 +0100 |
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| grillatzo@hotmail.com writes:
> Suppose you are truthfully told that ten marbles were inserted into
> a box, all of them identical except that their colors were
> determined by the toss of an unbiased coin. When heads came up, a
> white marble was inserted, and when tails came up, a black one. You
> reach into the box, draw out a marble, inspect its color, then
> return it to the box. You shake the box to mix the marbles randomly,
> and then reach in and again select a marble at random. If you
> inspect ten marbles in succession in this manner and all turn out to
> be white, what is the probability to the nearest whole percent that
> all ten marbles in the box are white?
What mathematic competition is this from? Anyway, here is how to
tackle this easily enough:
If you have n white marbles in the box, the probability of getting k
white draws from 10 is obviously
(n/10)^k * (1-(n/10))^(10-k) * choose(10,k).
The a-priori probability of getting n marbles is (also obviously)
choose(10,n)/1024
Now you just need to apply Bayes.
I get
p(n in box|k drawn) = p(k drawn|n in box)*p(n in box)/p(k drawn)
10
( )
n k n 10 - k 10 n
(--) (1 - --) ( ) * ----
10 10 k 1024
= -------------------------------------
10 10
---- ( )
\ n k n 10 - k 10 n
> (--) (1 - --) ( ) * ----
/ 10 10 k 1024
----
n = 0
Tabulating this with k downwards and n across (notice that n is a free
variable only in the numerator), I get
[ [ 0.070193, 0.24474, 0.33916, 0.23793, 0.08913, 0.017274, 1.5456e-3, 4.9737e-5, 3.2345e-7, 7.019e-11, 0. ]
[ 0., 0.092786, 0.28931, 0.34794, 0.20274, 0.058939, 7.9106e-3, 3.9598e-4, 4.4144e-6, 2.1555e-9, 0. ]
[ 0., 0.023503, 0.16488, 0.33992, 0.30813, 0.13437, 0.027051, 2.1063e-3, 4.0254e-5, 4.4224e-8, 0. ]
[ 0., 4.5418e-3, 0.071694, 0.25339, 0.35728, 0.2337, 0.070574, 8.5484e-3, 2.8005e-4, 6.9225e-7, 0. ]
[ 0., 6.9529e-4, 0.024694, 0.14961, 0.32816, 0.32197, 0.14585, 0.027479, 1.5434e-3, 8.5838e-6, 0. ]
[ 0., 8.5815e-5, 6.8574e-3, 0.071225, 0.24301, 0.35764, 0.24301, 0.071225, 6.8574e-3, 8.5815e-5, 0. ]
[ 0., 8.5838e-6, 1.5434e-3, 0.027479, 0.14585, 0.32197, 0.32816, 0.14961, 0.024694, 6.9529e-4, 0. ]
[ 0., 6.9225e-7, 2.8005e-4, 8.5484e-3, 0.070574, 0.2337, 0.35728, 0.25339, 0.071694, 4.5418e-3, 0. ]
[ 0., 4.4224e-8, 4.0254e-5, 2.1063e-3, 0.027051, 0.13437, 0.30813, 0.33992, 0.16488, 0.023503, 0. ]
[ 0., 2.1555e-9, 4.4144e-6, 3.9598e-4, 7.9106e-3, 0.058939, 0.20274, 0.34794, 0.28931, 0.092786, 0. ]
[ 0., 7.019e-11, 3.2345e-7, 4.9737e-5, 1.5456e-3, 0.017274, 0.08913, 0.23793, 0.33916, 0.24474, 0.070193 ] ]
So you would be interested in the lowest line (10 white marbles drawn)
and in there the last column (10 white marbles in the box). It is
somewhat funny that when drawing 10 white marbles, the most probable
contents of the box (about 1/3 probability) is 8 marbles, with 7 and 9
marbles being not too far behind (about 1/4 each), while 10 white
marbles in the box is not very likely (of course, 10 white marbles in
the box is completely impossible if I draw anything but 10 marbles,
nevertheless).
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
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| | From: | |-|erc | | Subject: | Re: PROBABILITIES | | Date: | Mon, 24 Jan 2005 16:02:33 +1000 |
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| "David Kastrup" wrote in
> grillatzo@hotmail.com writes:
>
> > Suppose you are truthfully told that ten marbles were inserted into
> > a box, all of them identical except that their colors were
> > determined by the toss of an unbiased coin. When heads came up, a
> > white marble was inserted, and when tails came up, a black one. You
> > reach into the box, draw out a marble, inspect its color, then
> > return it to the box. You shake the box to mix the marbles randomly,
> > and then reach in and again select a marble at random. If you
> > inspect ten marbles in succession in this manner and all turn out to
> > be white, what is the probability to the nearest whole percent that
> > all ten marbles in the box are white?
7%
Herc
>
> What mathematic competition is this from? Anyway, here is how to
> tackle this easily enough:
>
> If you have n white marbles in the box, the probability of getting k
> white draws from 10 is obviously
> (n/10)^k * (1-(n/10))^(10-k) * choose(10,k).
>
>
> The a-priori probability of getting n marbles is (also obviously)
>
> choose(10,n)/1024
>
> Now you just need to apply Bayes.
>
> I get
> p(n in box|k drawn) = p(k drawn|n in box)*p(n in box)/p(k drawn)
> 10
> ( )
> n k n 10 - k 10 n
> (--) (1 - --) ( ) * ----
> 10 10 k 1024
> = -------------------------------------
> 10 10
> ---- ( )
> \ n k n 10 - k 10 n
> > (--) (1 - --) ( ) * ----
> / 10 10 k 1024
> ----
> n = 0
>
> Tabulating this with k downwards and n across (notice that n is a free
> variable only in the numerator), I get
>
> [ [ 0.070193, 0.24474, 0.33916, 0.23793, 0.08913, 0.017274, 1.5456e-3, 4.9737e-5, 3.2345e-7, 7.019e-11, 0. ]
> [ 0., 0.092786, 0.28931, 0.34794, 0.20274, 0.058939, 7.9106e-3, 3.9598e-4, 4.4144e-6, 2.1555e-9, 0. ]
> [ 0., 0.023503, 0.16488, 0.33992, 0.30813, 0.13437, 0.027051, 2.1063e-3, 4.0254e-5, 4.4224e-8, 0. ]
> [ 0., 4.5418e-3, 0.071694, 0.25339, 0.35728, 0.2337, 0.070574, 8.5484e-3, 2.8005e-4, 6.9225e-7, 0. ]
> [ 0., 6.9529e-4, 0.024694, 0.14961, 0.32816, 0.32197, 0.14585, 0.027479, 1.5434e-3, 8.5838e-6, 0. ]
> [ 0., 8.5815e-5, 6.8574e-3, 0.071225, 0.24301, 0.35764, 0.24301, 0.071225, 6.8574e-3, 8.5815e-5, 0. ]
> [ 0., 8.5838e-6, 1.5434e-3, 0.027479, 0.14585, 0.32197, 0.32816, 0.14961, 0.024694, 6.9529e-4, 0. ]
> [ 0., 6.9225e-7, 2.8005e-4, 8.5484e-3, 0.070574, 0.2337, 0.35728, 0.25339, 0.071694, 4.5418e-3, 0. ]
> [ 0., 4.4224e-8, 4.0254e-5, 2.1063e-3, 0.027051, 0.13437, 0.30813, 0.33992, 0.16488, 0.023503, 0. ]
> [ 0., 2.1555e-9, 4.4144e-6, 3.9598e-4, 7.9106e-3, 0.058939, 0.20274, 0.34794, 0.28931, 0.092786, 0. ]
> [ 0., 7.019e-11, 3.2345e-7, 4.9737e-5, 1.5456e-3, 0.017274, 0.08913, 0.23793, 0.33916, 0.24474, 0.070193 ] ]
>
> So you would be interested in the lowest line (10 white marbles drawn)
> and in there the last column (10 white marbles in the box). It is
> somewhat funny that when drawing 10 white marbles, the most probable
> contents of the box (about 1/3 probability) is 8 marbles, with 7 and 9
> marbles being not too far behind (about 1/4 each), while 10 white
> marbles in the box is not very likely (of course, 10 white marbles in
> the box is completely impossible if I draw anything but 10 marbles,
> nevertheless).
>
>
> --
> David Kastrup, Kriemhildstr. 15, 44793 Bochum
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