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

The following problem is from both the 2010 AMC 12A #19 and 2010 AMC 10A #23, so both problems redirect to this page.

## 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 1

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

To stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \ldots, n-1,$ and draw a red marble from box $n.$ Thus, $$P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.$$

So, we must have $\frac{1}{n(n + 1)} < \frac{1}{2010}$ or $n(n+1) > 2010.$

Since $n(n+1)$ increases as $n$ increases, we can simply test values of $n$; after some trial and error, we get that the minimum value of $n$ is $\boxed{\textbf{(A) }45}$, since $45(46) = 2070$ but $44(45) = 1980.$

## Solution 2(cheap)

Do the same thing as Solution 1, but when we get to $n(n+1)>2010$ just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than $2010$. We get $45(46)=2070$, which is greater than $2010$, so we are done. The answer is $\textbf{(A)}$

-vsamc