Subject:
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Re: math question (or pattern... whatever...)
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Mon, 3 Mar 2003 23:30:44 GMT
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Viewed:
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277 times
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In lugnet.off-topic.geek, Ross Crawford writes:
<snip>
> Your friend has assumed that there is a difference between the reds, and you
> know which one you have seen in your hand. As there is no information in the
> original problem to substantiate this assumption, you have to assume it's
> not the case, so the valid combinations are in fact:
>
> r,r
> r,b
> r,w
> b,w
>
> Your friend rightly discounted the b,w combination, which leaves 3, of which
> yours must be one, therefore you are correct, the chances are 33%.
>
> ROSCO
I like driving home from work--gives me time to think about stuff...
And after reading all the great replies, I'm going to propose the following
solution:
If you have a bag with 4 marbles, 2 red, one white, one blue
and you reach your hand in and grab 2 of the marbles...
you have the following possibilities in your hand
r1 r2
r1 b
r1 w
r2 b
r2 w
b w
The second you expose one of the marbles in your hand and you know it's red,
you have the following:
r(exposed) r(unexposed)
r(e) b
r(e) w
r(u) b
r(u) w
b w
and, as shown by the separation, you say "If you know one is red (exposed)
what is the chance that the other one is red (unexposed)?"
the last 3 options are completely invalid, for they don't fit the criteria
"If one red is exposed..." (r(exposed))
so we're left with the first 3, and the chance that the r(unexposed) is
combined with r(exposed) is 1 in 3...
Thanks for all the help!
Dave K
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Message is in Reply To:
 | | Re: math question (or pattern... whatever...)
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| (...) Your friend has assumed that there is a difference between the reds, and you know which one you have seen in your hand. As there is no information in the original problem to substantiate this assumption, you have to assume it's not the case, (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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