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# Each of 2010 boxes in a line contains a single red marble, and for 1

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Math Expert
Joined: 02 Sep 2009
Posts: 54371
Each of 2010 boxes in a line contains a single red marble, and for 1  [#permalink]

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31 Mar 2019, 20:21
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Difficulty:

55% (hard)

Question Stats:

29% (02:17) correct 71% (03:26) wrong based on 7 sessions

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Each of 2010 boxes in a line contains a single red marble, and for $$1 \leq k \leq 2010$$, the box in the $$k_{th}$$ position also contains k white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let P(n) be the probability that Isabella stops after drawing exactly n marbles. What is the smallest value of n for which $$P(n) < \frac{1}{2010}$$?

A. 45
B. 63
C. 64
D. 201
E. 1005

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Re: Each of 2010 boxes in a line contains a single red marble, and for 1  [#permalink]

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31 Mar 2019, 20:48
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P(3) = white in first two 1/2*2/3= 1/3 and red in 3 rd = 1/3 hence 1/9.
Expanding it to p(45) the lowest option results in 1/45*1/45 = 1/2025 less than RHS hence option A

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Intern
Joined: 28 Jan 2019
Posts: 43
Each of 2010 boxes in a line contains a single red marble, and for 1  [#permalink]

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01 Apr 2019, 00:37
1
arvindmurugapan wrote:
P(3) = white in first two 1/2*2/3= 1/3 and red in 3 rd = 1/3 hence 1/9.
Expanding it to p(45) the lowest option results in 1/45*1/45 = 1/2025 less than RHS hence option A

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But shouldn't the probab of picking red in 3rd box be 1/4, as there are 3 white balls and 1 red ball?

Anyway, your answer still works, as $$P(n) = \frac{1}{n(n+1)}$$.
Therefore, substituting the least option 45 gives 1/2070. Thus A.
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Joined: 22 Nov 2018
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GMAT 1: 640 Q45 V35
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Re: Each of 2010 boxes in a line contains a single red marble, and for 1  [#permalink]

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01 Apr 2019, 00:49
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OhsostudiousMJ wrote:
arvindmurugapan wrote:
P(3) = white in first two 1/2*2/3= 1/3 and red in 3 rd = 1/3 hence 1/9.
Expanding it to p(45) the lowest option results in 1/45*1/45 = 1/2025 less than RHS hence option A

Posted from my mobile device

But shouldn't the probab of picking red in 3rd box be 1/4, as there are 3 white balls and 1 red ball?

Anyway, your answer still works, as $$P(n) = \frac{1}{n(n+1)}$$.
Therefore, substituting the least option 45 gives 1/2070. Thus A.

You are correct. P(3) will be 1/3*1/4
P(45)= 1/45*1/46= 1/2070 which is less than RHS.
P(44) = 1/44*1/45= 1980 which is greater than RHS.
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Re: Each of 2010 boxes in a line contains a single red marble, and for 1   [#permalink] 01 Apr 2019, 00:49
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