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

2010 AMC 12A (Problems • Answer Key • Resources)
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2010 AMC 12A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2(cheap)

See also