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 10A Problems/Problem 23

Revision as of 23:13, 2 February 2012 by Flamewire (talk | contribs) (→‎Solution)

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

Solution 1

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 $\boxed{45\ \textbf{(A)}}$.

Solution 2

Using the first few values of $n$, it is easy to derive a formula for $P(n)$. The chance that she stops on the second box ($n=2$) is the chance of drawing a white marble then a red marble: $\frac{1}2 * \frac{1}3$. The chance that she stops on the third box ($n=3$) is the chance of drawing two white marbles then a red marble:$\frac{1}2 * \frac{2}3 * \frac{1}4$. If $n=4$, $P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5$.

Cross-cancelling in the fractions gives $P(2) =$\frac{1}{2*3}$,$P(3) = $\frac{1}{3*4}%, and$ (Error compiling LaTeX. Unknown error_msg)P(4) = $\frac{1}{4*5}$. From this, it is clear that $P(n) = \frac{1}{(n)(n+1)}$. (Alternatively, $P(n) = \frac{(n-1)!}{(n+1)!}$.)

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

The lowest integer that satisfies the above inequality is 45.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_23&oldid=44441"