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

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Intern
Joined: 26 Aug 2013
Posts: 11
There is a set of beads, each of which is painted either red  [#permalink]

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30 Aug 2013, 01:06
2
11
00:00

Difficulty:

85% (hard)

Question Stats:

49% (02:17) correct 51% (01:48) wrong based on 251 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.

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?
Intern
Joined: 31 Jan 2013
Posts: 17
Schools: ISB '15
WE: Consulting (Energy and Utilities)
Re: probability question from Princeton review  [#permalink]

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30 Aug 2013, 02:12
1
1
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
Intern
Joined: 07 Dec 2016
Posts: 40
Re: There is a set of beads, each of which is painted either red  [#permalink]

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16 Jun 2017, 02:07
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.
Senior Manager
Joined: 02 Apr 2014
Posts: 474
GMAT 1: 700 Q50 V34
There is a set of beads, each of which is painted either red  [#permalink]

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21 Nov 2017, 11:23
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

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

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Joined: 09 Sep 2013
Posts: 9465
Re: There is a set of beads, each of which is painted either red  [#permalink]

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18 Dec 2018, 00:58
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Re: There is a set of beads, each of which is painted either red &nbs [#permalink] 18 Dec 2018, 00:58
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