Subject:
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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 20:54:01 GMT
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Viewed:
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247 times
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k, here's something that came across my desk...
you have a bag with 4 marbles in it.
2 of the marbles are red
1 marble is white
1 marble is blue
you reach in and grab 2 random marbles
you open up your hand a bit to only see the colour of one of those 2
marbles, and it happens to be red
what is the chance that the other marble in your hand is also red?
I say it's 33 percent
my friend says it's 20 percent
I worked it out like this:
4 marbles:
r r b w
you grab 2, one is red so you immediately know that one of the red's doesn't
have to be counted.
so you're left with
r b w
that may be in your hand
so one in 3 = 33 percent
My friend calculated it this way
r(1) r(2) b w
the 'event' is that you have 2 balls in your hand, one of which is red
so out of all the possible unique combinations (6)--
r(1) r(2)
r(1) b
r(1) w
r(2) b
r(2) w
b w
the last one (b w) is eliminated 'cause you already know that there is a red
in your hand, so the last one isn't part of 'the event'
so the possibility of the second marble being red is 1 in 5, or 20 percent.
but what my friend won't admit to, is that if you have
r(1) r(2)
r(1) b
r(1) w
r(2) b
r(2) w
(b w being still eliminated due to red already being in your hand) that a
last possible combination:
r(2) r(1)
fits into the mix, so 2 out of 6, which still equals 33 percent
Any stats people out there that could shed logic on this?
Thanks
Dave K
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Message has 5 Replies:
 | | Re: math question (or pattern... whatever...)
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| (...) <long snip> you're right, in both senses. Your logic for the first solution is valid, and your identification of his error in the second solution (which would lead to the same answer) is also correct. It is, indeed, 33% Adrian -- (URL) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
| | | Re: math question (or pattern... whatever...)
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| (...) I'm not a stats person, but I agree with your friend. r(2) r(1) is the same as r(1) r(2), because order doesn't matter. If it does, then you end up with more combos (w r(2), etc)... -- that's what I was going to say. But, I decided to run a (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
| | | Re: math question (or pattern... whatever...)
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| (...) Yep, you're right, your friend is wrong. The problem is that you *wouldn't* count "r(2) r(1)" if order didn't matter. But then again, the thing he's forgetting is that if you don't count "r(2) r(1)", then you also CAN'T count *both* "r(1) b" (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
| | | 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)
| | | Re: math question (or pattern... whatever...)
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| (...) I think the two answers depend on a more accurate statement of the problem. If the problem is stated that you pick two marbles, and given that the first one you pick is red, what is the chance the second one is red, then order matters, and the (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
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