Difference between revisions of "2010 AMC 10A Problems/Problem 23"

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== Problem ==
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#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>?
 
 
 
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math>
 
 
 
== Solution ==
 
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>.
 
 
 
The probability of drawing a red marble at box <math>n</math> is therefore
 
 
 
<cmath>\begin{align*}\frac{1}{n+1} \left( \prod_{k&=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}\\
 
\frac{1}{n+1} \left( \frac{1}{n} \right) &< \frac{1}{2010}\\
 
(n+1)n &> 2010\end{align*}</cmath>
 
 
 
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>.
 
 
 
== See also ==
 
{{AMC10 box|year=2010|num-b=22|num-a=24|ab=A}}
 
 
 
[[Category:Introductory Combinatorics Problems]]
 

Latest revision as of 13:28, 26 May 2020

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