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

(Created page with '== Problem == 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 …')
 
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The probability of drawing a red marble at box <math>n</math> is therefore
 
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}\\
+
<center>
−
\frac{1}{n+1} \left( \frac{1}{n} \right) &< \frac{1}{2010}\\
+
<math>\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}</math>
−
(n+1)n &> 2010\end{align*}</cmath>
+
 
 +
<math>\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}</math>
 +
 
 +
<math>(n+1)n > 2010</math>
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</center>
  
 
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>.
 
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>.

Revision as of 14:03, 2 April 2010

Problem

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$, the box in the $k\text{th}$ 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 $P(n) < \frac{1}{2010}$?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$

Solution

The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$. The probability of drawing a red marble from box $n$ is $\frac{1}{n+1}$.

The probability of drawing a red marble at box $n$ is therefore

$\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}$

$\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}$

$(n+1)n > 2010$

It is then easy to see that the lowest integer value of $n$ that satisfies the inequality is $\boxed{45\ \textbf{(A)}}$.

See also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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