2010 AMC 12A Problems/Problem 19: Difference between revisions

Revision as of 22:21, 11 February 2010

Problem 19

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

Revision as of 20:47, 10 February 2010 view source Stargroup (talk \| contribs) 28 edits m →Solution ← Older edit		Revision as of 22:21, 11 February 2010 view source Mathgeek2006 (talk \| contribs) editor 430 edits m →Solution Newer edit →
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	It is then easy to see that the lowest integer value of <math>n</math> that satisfies the inequality is <math>\boxed{45\ \textbf{(D)}}</math>.		It is then easy to see that the lowest integer value of <math>n</math> that satisfies the inequality is <math>\boxed{45\ \textbf{(A)}}</math>.

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2010 AMC 12A Problems/Problem 19: Difference between revisions

Revision as of 22:21, 11 February 2010

Problem 19

Solution