2010 Boxes of Marbles

Each of 2010 2010 boxes in a line contains one red marble, and for 1 k 2010 1\leq k\leq2010 , the box in the k t h k^{th} position also contains k white marbles. A child begins at the first box and successively draws a single marble at random from each box in order. He stops when he first draws a red marble. Let P(n) be the probability that he stops after drawing exactly n marbles. The possible values(s) of n for which P ( n ) < 1 2010 P(n)<\frac{1}{2010}

25,43,56 37,53,64 45,70,98 44,54,65

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1 solution

Stephen Brown
Dec 14, 2017

The probability of drawing exactly n n marbles can be calculated as the product of drawing n 1 n-1 white marbles followed by a red marble:

P ( n ) = 1 2 × 2 3 × × n 1 n × 1 n + 1 = 1 n ( n + 1 ) P(n) = \frac{1}{2}\times\frac{2}{3}\times\dots\times\frac{n-1}{n}\times\frac{1}{n+1} = \frac{1}{n(n+1)}

Solving P ( n ) < 1 2010 P(n) < \frac{1}{2010} gives us

n ( n + 1 ) > 2010 n 45 n(n+1) > 2010 \Rightarrow \boxed{n\geq45}

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