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31 Mar 2019, 20:21
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
43% (02:23) correct
57%
(03:23)
wrong
based on 14
sessions
History
Date
Time
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Not Attempted Yet
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
A. 45
B. 63
C. 64
D. 201
E. 1005
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
Posted from my mobile device
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
OhsostudiousMJ
Joined: 28 Jan 2019
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Schools: ISB '21 (A) Great Lakes '21 (WD)
GMAT 1: 700 Q49 V36
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01 Apr 2019, 00:37
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arvindmurugapan
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.
