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| == Problem ==
| | #REDIRECT [[2010_AMC_12A_Problems/Problem_19]] |
| 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>?
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| <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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| == Solutions ==
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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 red marble at box <math>n</math> is therefore
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| <center>
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| <math>\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}</math>
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| <math>\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}</math>
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| <math>(n+1)n > 2010</math>
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| </center>
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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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| An easy way to know that <math>45</math> is the answer is that <math>50*51=2550</math>, so you know <math>n<50</math> - the only solution for n under <math>50</math> is <math>45</math>.
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| === Solution 2 ===
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| Using the first few values of <math>n</math>, it is easy to derive a formula for <math>P(n)</math>. The chance that she stops on the second box (<math>n=2</math>) is the chance of drawing a white marble then a red marble: <math>\frac{1}2 \cdot \frac{1}3</math>. The chance that she stops on the third box (<math>n=3</math>) is the chance of drawing two white marbles then a red marble:<math>\frac{1}2 \cdot \frac{2}3 \cdot \frac{1}4</math>. If <math>n=4</math>, <math>P(n) = \frac{1}2 \cdot \frac{2}3 \cdot \frac{3}4 \cdot \frac{1}5</math>.
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| Cross-cancelling in the fractions gives <math>P(2) =\frac{1}{2\cdot3}</math>, <math>P(3) = \frac{1}{3\cdot4}</math>, and <math>P(4) = \frac{1}{4\cdot5}</math>. From this, it is clear that <math>P(n) = \frac{1}{(n)(n+1)}</math>. (Alternatively, <math>P(n) = \frac{(n-1)!}{(n+1)!}</math>.)
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| <math>\frac{1}{(n+1)(n)} < \frac{1}{2010}</math>
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| The lowest integer that satisfies the above inequality is <math>\boxed{(A) 45}</math>.
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| == See also ==
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| {{AMC10 box|year=2010|num-b=22|num-a=24|ab=A}}
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| [[Category:Introductory Combinatorics Problems]] | |
| {{MAA Notice}}
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