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Difference between revisions of "2010 AMC 12A Problems/Problem 19"

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{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #19]] and [[2010 AMC 10A Problems|2010 AMC 10A #23]]}}
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== Problem ==
 
== Problem ==
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Each of 2010 boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> 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 <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?
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Each of <math>2010</math> boxes in a line contains a single red marble, and for <math>1 \le k \le 2010</math>, the box in the <math>k\text{th}</math> position also contains <math>k</math> 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 <math>P(n)</math> be the probability that Isabella stops after drawing exactly <math>n</math> marbles. What is the smallest value of <math>n</math> for which <math>P(n) < \frac{1}{2010}</math>?
  
 
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math>
 
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math>
  
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== Solution ==
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== Solution 1==
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The probability of drawing a white marble from box <math>k</math> is <math>\frac{k}{k+1}</math>. The probability of drawing a red marble from box <math>n</math> is <math>\frac{1}{n+1}</math>.
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The probability of drawing a white marble from box <math>k</math> is <math>\frac{k}{k + 1}</math>, and the probability of drawing a red marble from box <math>k</math> is <math>\frac{1}{k+1}</math>.
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To stop after drawing <math>n</math> marbles, we must draw a white marble from boxes <math>1, 2, \ldots, n-1,</math> and draw a red marble from box <math>n.</math> Thus, <cmath>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)}.</cmath>
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So, we must have <math>\frac{1}{n(n + 1)} < \frac{1}{2010}</math> or <math>n(n+1) > 2010.</math>
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Since <math>n(n+1)</math> increases as <math>n</math> increases, we can simply test values of <math>n</math>; after some trial and error, we get that the minimum value of <math>n</math> is <math>\boxed{\textbf{(A) }45}</math>, since <math>45(46) = 2070</math> but <math>44(45) = 1980.</math>
  
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The probability of drawing a red marble at box <math>n</math> is therefore
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== Solution 2(cheap) ==
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Do the same thing as Solution 1, but when we get to <math>n(n+1)>2010</math> just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than <math>2010</math>. We get <math>45(46)=2070</math>, which is greater than <math>2010</math>, so we are done. The answer is <math>\textbf{(A)}</math>
  
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<cmath>\begin{align*}\frac{1}{n+1} \left( \prod_{k&=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}\\
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-vsamc
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\frac{1}{n+1} \left( \frac{1}{n} \right) &< \frac{1}{2010}\\
 
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(n+1)n &> 2010\end{align*}</cmath>
 
  
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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{(A)}}</math>.
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==Video Solution by TheBeautyofMath==
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https://www.youtube.com/watch?v=47XsxmQ5Ej4
  
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== See also ==
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== See Also ==
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{{AMC10 box|year=2010|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}}
 
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 17:30, 10 July 2022

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

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=47XsxmQ5Ej4

See Also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions
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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