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

2010 AMC 12A Problems/Problem 19

Revision as of 16:10, 10 February 2010 by Stargroup (talk | contribs) (Created page with '== Problem 19 == Each of 2010 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 <…')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 19

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


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


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


$(n+1)n > 2010$


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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_19&oldid=33390"