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

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

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New post 17 Jan 2020, 01:16
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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.


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Re: There is a set of beads, each of which is painted either red or blue.  [#permalink]

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New post 17 Jan 2020, 07:47
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Interesting setup. If we just care whether we had more red than blue beads to start with, the 'merged' beads don't really matter, because half of them were originally red, and half were originally blue, so they even out. Algebraically, if we end up with r red beads, b blue beads, and p merged beads, then we originally had r + (p/2) red beads, and b + (p/2) blue beads, and we will have started with more red than blue if

r + (p/2) > b + (p/2)

is true, and subtracting p/2 from both sides, we just need to know if r > b, or if we ended up with more pure red than pure blue beads.

Statement 1 tells us r < p + b. That's not sufficient; maybe p = 0, and r < b, but maybe b=0 and p is large, and r > b.

Statement 2 tells us b > p + r. But p + r > r, so b > r, which means we started with more blue than red beads, which is sufficient (we know the answer to the question is 'no'). So the answer is B.
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Re: There is a set of beads, each of which is painted either red or blue.   [#permalink] 17 Jan 2020, 07:47
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