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NickH
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Two bags of marbles
« on: Dec 9th, 2004, 3:29pm » |
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You have two bags of marbles. One bag contains two purple marbles and one orange marble, the other contains three purple marbles and one orange marble.
You select a bag at random and, from that bag, draw a marble at random. It is purple. You then draw a second marble at random from the same bag. What is the probability that this marble will be purple?
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Sir Col
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Re: Two bags of marbles
« Reply #1 on: Dec 10th, 2004, 9:56am » |
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Let p be probability of selecting 1st bag.
P(1st P AND 2nd P)=p(2/3)(1/2)+(1-p)(3/4)(2/3)=(3-p)/6
P(1st P)=p(2/3)+(1-p)(3/4)=(9-p)/12
P(2nd P|1st P)=P(1st P AND 2nd P)/P(1st P)=2(3-p)/(9-p)
Assuming that p=1/2, P(2nd P|1st P)=10/17.
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Nice problem, NickH, as it contains so many subtleties. If you don't mind, I'll borrow it for my sixth form, especially with this rather challenging extension that leads to a satisfying generalisation...
The first bag contains two purple and one orange marble, and the second bag contains more than one orange marble. Given that the first marble is purple, what is the smallest number of purple marbles in the second bag such that the probability of the second marble being purple is 1/2?
(Assume the probability of selecting either bag is 1/2)
What if the probability of the second marble being purple is q?
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NickH
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Re: Two bags of marbles (plus new puzzle)
« Reply #2 on: Dec 10th, 2004, 12:33pm » |
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Feel free to borrow it for your sixth form, Sir Col!
I used a slightly different approach. For the first draw, imagine drawing each marble from the first bag 4 times and each marble from the second bag 3 times. This gives us 12 drawings from each bag, necessary as each bag must be equally likely to be selected.
As we have drawn 8 purple marbles from the first bag and 9 purple marbles from the second bag, the probability of having selected the first bag, given that the first draw is purple, is 8/17.
Hence p(2nd P|1st P) = (8/17)(1/2) + (9/17)(2/3) = 10/17.
It's a long story, but my invention of this puzzle was triggered by a discussion of the ambiguities of the following (much better) puzzle!
The Loose Marbles Company manufactures orange marbles and purple marbles. A bag of their marbles may contain any combination of orange and purple marbles (including no orange marbles and including no purple marbles), and all combinations are equally probable.
A boy bought a bag of their marbles and pulled out one at random without looking. Given that it was purple, what is the probability that if he pulled out a second marble at random it would also be purple?
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rmsgrey
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Re: Two bags of marbles (plus new puzzle)
« Reply #3 on: Dec 10th, 2004, 12:43pm » |
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on Dec 10th, 2004, 12:33pm, NickH wrote:The Loose Marbles Company manufactures orange marbles and purple marbles. A bag of their marbles may contain any combination of orange and purple marbles (including no orange marbles and including no purple marbles), and all combinations are equally probable.
A boy bought a bag of their marbles and pulled out one at random without looking. Given that it was purple, what is the probability that if he pulled out a second marble at random it would also be purple? |
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Since all combinations are equally probable, the average bag must be rather large...
With that in mind, I'll say ::0.5 ::
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NickH
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Re: Two bags of marbles
« Reply #4 on: Dec 10th, 2004, 12:55pm » |
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Quote:| Since all combinations are equally probable, the average bag must be rather large... |
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Indeed -- that is one of the ambiguities!
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Sir Col
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Re: Two bags of marbles
« Reply #5 on: Dec 10th, 2004, 3:58pm » |
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On the one hand, the meaning of the statement, "all combinations are equally probable," does not seem clear. As the size of the bag is unknown, the strings representing combinations: OBBOOBO... are unbounded. So we may ask the question, what is actually knowable in this situation?
However, if we examine the of outcomes if the size of the sample were known...
2 marbles: OO, OP, PO, PP; P(2nd P|1st P)=1/2
3 marbles: OOO, OOP, OPO, OPP, POO, POP, PPO, PPP; P(2nd P|1st P)=2/4=1/2.
It appears that the answer is always 1/2, and, of course, in general we note that, P(2nd P|1st P)=P(1st P AND 2nd P)/P(1st P)=(1/2)(1/2)/(1/2)=1/2.
In fact, it appears to be one of the variations of the mother and two children riddle (see #2); as the outcome is independent, P(2nd P)=1/2.
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NickH
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Re: Two bags of marbles
« Reply #6 on: Dec 10th, 2004, 4:31pm » |
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Sir Col, what would you think if I suggested that one possible interpretation of "all combinations are equally probable" is that, given a specific number of marbles in the bag, say, n, then the probability that 0 <= a <= n of them are purple is simply 1/(n+1)?
This rather unconventional interpretation of the word 'combination' is based upon the fact that the marbles in the bag are presumably indistinguishable, except by colour.
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Sir Col
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Re: Two bags of marbles
« Reply #7 on: Dec 11th, 2004, 2:54am » |
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Interestingly it reminded me of a "discussion" I had with my sixth form when I presented them with a variation of #2 in the post I linked to above. I'll post the version from this forum:
You are at the park, pushing your daughter on a swing. You meet a woman with her daughter and start a conversation. She says she has two children. What is the chance that both children are girls?
First interpretation:
If the probability of any given child being boy or girl is 1/2, then independent of the child present, the probability of the other child being a girl is 1/2. So P(GG)=1/2.
Second interpretation:
As one child is known to be a girl, the possible combinations (listed in order of age) are BG, GB, GG. As each are equally likely, P(GG)=1/3.
It seems that your suggestion is a valid interpretation and what this would produce is a rectanglular distribution of the number of purples in a bag:
Let X be number of purples in bag of size n, and P(X=r)=1/(n+1).
P(1st P AND 2nd P)=P(X=2)*P(PP|X=2)+P(X=3)*P(PP|X=3)+...+P(X=n)*P(PP|X=n)
=(1/(n+1))[(2/n)(1/(n-1))+(3/n)(2/(n-1))+(4/n)(3/(n-1))+...+(n/n)((n-1)/(n-1))]
=(1/(n(n+1)(n-1))[2*1+3*2+4*3+...+n(n-1)]
=(2/(n(n+1)(n-1))[2*1/2+3*2/2+...+n(n-1)/2]
As the sum of n triangle numbers is n(n+1)(n+2)/6
P(1st P AND 2nd P)=(2/(n(n+1)(n-1))(n(n+1)(n-1))/6=1/3 - weird!
P(1st P)=P(X=1)*P(P|X=1)+P(X=2)*P(P|X=2)+...+P(X=n)*P(P|X=n)
=(1/(n+1))[1/n+2/n+3/n+...+n/n]
=(1/(n(n+1)))[1+2+...+n]
=(1/(n(n+1)))(n(n+1)/2)=1/2
Therefore, P(2nd P|1st P)=(1/3)/(1/2)=2/3, which is entirely independent of n. Unless I've made a terrible oversight, this is intuitively disturbing!
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NickH
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Re: Two bags of marbles
« Reply #8 on: Dec 11th, 2004, 9:23am » |
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It is indeed interesting that the answer, given those assumptions, is independent of n!
Re the two interpretations of the conversation in the park, this goes to the heart of the concept of ordered pair. Did you manage to convince everyone in your class that the second interpretation is incorrect?!
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Sir Col
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Re: Two bags of marbles
« Reply #9 on: Dec 11th, 2004, 7:07pm » |
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I did in the end, but it required a a good deal of persuasion. I argued...
There are four possibilities. The girl present is the:
(1) eldest and the youngest is a girl.
(2) eldest and the youngest is a boy.
(3) youngest and the eldest is a girl.
(4) youngest and the eldest is a boy.
Therefore P(two girls)=2/4=1/2
It was a case of trying to convince them that if they are going to list possible arrangements with some ordering principle (age in this case), then although BG and GB represent two possiblilities, GG by itself does not entirely represent the alternatives. That is, if the mother has two daughters and we are looking at one of them, she could be the eldest or the youngest, thus representing two uniquely different possibilities.
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NickH
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Re: Two bags of marbles
« Reply #10 on: Dec 12th, 2004, 5:28am » |
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Or you could adopt a different ordering principle: the order in which we meet her two children! This is more immediately applicable, as we don't know the relative ages of her children. We then get possible ordered pairs GG, GB, giving p(two girls) = 1/2.
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Sir Col
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Re: Two bags of marbles
« Reply #11 on: Dec 12th, 2004, 6:54am » |
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I tried that initially, but they were not having any of it after convincing them that the answer to...
You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. She says yes. What is the chance that both children are girls?
... was 1/3, because they could be BG, GB, or GG. They simply couldn't see the difference; there wasn't a "method" to work it out. Whereas with the age ordering principle they've seen it before and it is "safe". After going through problems like this, the usual responses are, "Do we need to know this for the exam?" or "Can you give us more examples like this so I can practice them?" *boo-hoo* 
Perhaps we were no different when we were their age? It's funny that they seem to spend all of their time trying to find problems that are related and similar to ones they've seen before. Yet we spend all of our time trying to find problems that are different to anything we've seen before.
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Icarus
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Re: Two bags of marbles
« Reply #12 on: Dec 12th, 2004, 10:59am » |
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We were exactly like that when we were their age. They know that they are expected to produce solutions to problems. Therefore a problem that they can't solve becomes a threat. But in this setting, we know that failure to find a solution is no big deal. Someone else can solve it. And even if no one does, it has no negative impact on our lives (beyond the impact we allow it have on our own psyche).
When you work for grades, new stuff threatens. When you work for fun, new stuff challenges.
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towr
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Re: Two bags of marbles
« Reply #13 on: Dec 12th, 2004, 11:44am » |
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on Dec 12th, 2004, 10:59am, Icarus wrote:| When you work for grades, new stuff threatens. When you work for fun, new stuff challenges. |
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Am I ever glad I never worked for grades
I suppose that must have not made me like them at that age..
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