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2010 AMC 12A Problems/Problem 19

Revision as of 22:26, 16 February 2014 by PythagorasofSamos (talk | contribs) (→‎Solution 2)

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

The probability of drawing a white marble from the first box is $\frac{1}{2}$. The probability of drawing a white marble from the second box is $\frac{2}{3}$.

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

From this, we find that \[P(n) = (\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \dots \cdot \frac {n - 1}{n}) \cdot \frac{1}{n +1}\]

Clearly, \[P(n) = \frac{(n - 1)!}{(n + 1)!} = \frac{1}{n (n + 1)}\]

So \[\frac{1}{n(n + 1)} < \frac{1}{2010}\] \[n(n+1) > 2010\]

The minimum integer value of $n$ is obviously $\boxed{\textbf{(A)}45}$.

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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