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There is a set of beads, each of which is painted either red

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There is a set of beads, each of which is painted either red  [#permalink]

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New post 30 Aug 2013, 02:06
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Question Stats:

53% (02:23) correct 47% (02:26) wrong based on 259 sessions

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There is a set of beads, each of which is painted either red or blue. The beads are split, with each bead being cut in half. Then they are merged, where all the halves are randomly reassembled back into the same number of beads that there were to begin with. This results in r red beads, b blue beads, and p beads that are half red and half blue. Before the split and merge, were there more red beads than blue beads?

(1) after the split and merge, the probability of picking a bead that is only red is less than the probability of picking a bead that is at least half blue.
(2) After the split and merge, the probability of picking a bead that is only blue is greater than the probability of picking a bead that is at least half red.

Answer is B.
Well, I got the answer but it took me so long to solve it.
Can someone help me to solve this kind of questions quicker?
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Re: probability question from Princeton review  [#permalink]

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New post 30 Aug 2013, 03:12
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Consider the Number of Red and Blue beads as R and B respectively and, after splitting, the number of beads as R1,B1 and P(mixed)

We need to prove R>B
From the given information, We could say
R+B = R1+B1+P
Or
R = R1+P/2
B = B1+P/2

1) Given R1<B1+P -
R1<B+P/2 or R1+P/2<B+P
R<B+P - from this equation, there is no way we can interpret about R or B. Hence not Sufficient

2) Given B1>R1+P -
From this we could definitely say B>B1>R1+P
B> (R1+P/2)+P/2
So we can definitely say,
B>R
hence St2 is sufficent to answer the question.

I hope this helps
/SW
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Re: There is a set of beads, each of which is painted either red  [#permalink]

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New post 16 Jun 2017, 03:07
I got B right away as the only way the prob of only Blue beads > split Red ones is if there were more blue than red beads.
But can someone help do it algebraically as I'd like to see it efficiently laid out using algebra. Thank you.
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There is a set of beads, each of which is painted either red  [#permalink]

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New post 21 Nov 2017, 12:23
Let R number of full read beads before merge
Let B number of full blue beads before merge

Let r number of completely red beads after merge
Let b number of completely blue beads after merge
Let p number of half-read && half-blue beads after merge

R = (2r + p)/2
B = (2b + p)/2

Question is, before merge, R > B?
=> \((2r + p)/2 > (2b + p)/2\) ?
=> r > b ?


Statement (1) after the split and merge, the probability of picking a bead that is only red is less than the probability of picking a bead that is at least half blue.

probability of picking a bead that is only red = \(r / (r + b + p)\)
probability of picking a bead that is at least half blue = probalility of picking full blue OR probablity of half-blue = \((b/(r+b+p)) + (p/(r+b+p))\)

=> \(r / (r + b + p)\) < \((b/(r+b+p)) + (p/(r+b+p))\)
=> r < b + p => r can be less than or greater than b => Not suff

Statement (2) After the split and merge, the probability of picking a bead that is only blue is greater than the probability of picking a bead that is at least half red.

probability of picking a bead that is only blue = \(b / (r + b + p)\)
probability of picking a bead that is at least half red = probalility of picking full red OR probablity of half-red = \((r/(r+b+p)) + (p/(r+b+p))\)

=> \(b / (r + b + p)\) > \((r/(r+b+p)) + (p/(r+b+p))\)
=> b > r + p => b is definitely greater than r => Sufficient to say there were less Red beads than blue beads before merge

Answer (B)
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Re: There is a set of beads, each of which is painted either red  [#permalink]

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New post 18 Dec 2018, 01:58
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Re: There is a set of beads, each of which is painted either red   [#permalink] 18 Dec 2018, 01:58
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