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2010 AMC 10A Problems/Problem 23

Revision as of 14:03, 2 April 2010 by Moplam (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

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

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions
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