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| − | == Problem ==
| + | #REDIRECT [[2010_AMC_12A_Problems/Problem_19]] |
| − | Each of <math>2010</math> boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> 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 <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?
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| − | <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math>
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| − | == Solution ==
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| − | The probability of drawing a white marble from box <math>k</math> is <math>\frac{k}{k+1}</math>. The probability of drawing a red marble from box <math>n</math> is <math>\frac{1}{n+1}</math>.
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| − | The probability of drawing a red marble at box <math>n</math> is therefore
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| − | <cmath>\begin{align*}\frac{1}{n+1} \left( \prod_{k&=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}\\
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| − | \frac{1}{n+1} \left( \frac{1}{n} \right) &< \frac{1}{2010}\\
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| − | (n+1)n &> 2010\end{align*}</cmath>
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| − | It is then easy to see that the lowest integer value of <math>n</math> that satisfies the inequality is <math>\boxed{45\ \textbf{(A)}}</math>.
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| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=22|num-a=24|ab=A}}
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| − | [[Category:Introductory Combinatorics Problems]] | |
