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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Tue, 4 Mar 2003 00:50:53 GMT
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Viewed:
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250 times
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David Koudys wrote:
>
> 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
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 possible orders are:
r1 r2
r1 w
r1 b
r2 r1
r2 w
r2 b
w r1
w r2
w b
b r1
b r2
b w
Of these, only 6 meet the criteria that the 1st marble picked is red,
and therefore the probability of the 2nd one being red is 1 in 3.
Another statement of the problem is pick two marbles and show them to
me. I will tell you to throw them back if neither are red. If one is
red, I'll show it to you. What's the probability that the other one is
also red? Now order of picking doesn't matter, so out of the above
choices, there are 10 that qualify. Of these, only two have two red
marbles and thus the chance of the 2nd one being red is 1 in 5.
A third statement is pick two marbles. One in each hand. Pick a hand to
peek at. If it doesn't hold a red marble, throw them back. If it does,
what is the chance that the other marble is red. This is the same as the
"order matters" version, but some percentage of the time, you reveal
marble two instead of marble one. There are also 6 times when marble 2
is red of which 2 have marble 1 as red, and therefore a 1 in 3. This
problem is also the same as pick two marbles from the bag, put them into
another bag, then pick one of those, if it's red, what is the
probability the 2nd is red.
Frank
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Message is in Reply To:
 | | math question (or pattern... whatever...)
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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 (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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