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

## 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 $P(4) = \frac{1}{4*5}$. 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 $\boxed{(A) 45}$.