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

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Intern
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Joined: 26 Aug 2013
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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:

50% (02:49) correct 50% (01:32) wrong based on 157 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.

[Reveal] Spoiler:
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?
[Reveal] Spoiler: OA
1 KUDOS received
Intern
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Joined: 31 Jan 2013
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Schools: ISB '15
WE: Consulting (Energy and Utilities)
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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 02 Jun 2014, 08:53
I don't get it. Bunuel could you solve alternatively?

Question is to express half/half red and blue balls algebraically

Thanks!
Cheers
J :)
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Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
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Kudos [?]: 462 [0], given: 355

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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 02 Jun 2014, 08:53
I don't get it. Bunuel could you solve alternatively?

Question is to express half/half red and blue balls algebraically

Thanks!
Cheers
J :)
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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 02 Mar 2016, 05:17
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: There is a set of beads, each of which is painted either red   [#permalink] 02 Mar 2016, 05:17
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There is a set of beads, each of which is painted either red

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