Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 12A Problems/Problem 19"

m (Semi-automated contest formatting - script by azjps)
m (→‎Solution)
Line 9: Line 9:
 
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
  
−
<cmath>\begin{align*}\frac{1}{n+1} \left( \prod_{k&=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}\\
+
<cmath>\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}\\
+
\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}\\
−
(n+1)n &> 2010\end{align*}</cmath>
+
(n+1)n > 2010</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>.
 
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 14:01, 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 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 12 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_19&oldid=34166"