|
|
| (11 intermediate revisions by 9 users not shown) |
| Line 1: |
Line 1: |
| − | == 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>?
| |
| − | | |
| − | <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math>
| |
| − | | |
| − | == Solution ==
| |
| − | === Solution 1 ===
| |
| − | 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>.
| |
| − | | |
| − | The probability of drawing a red marble at box <math>n</math> is therefore
| |
| − | | |
| − | <center>
| |
| − | <math>\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}</math>
| |
| − | | |
| − | <math>\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}</math>
| |
| − | | |
| − | <math>(n+1)n > 2010</math>
| |
| − | </center>
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | === Solution 2 ===
| |
| − | | |
| − | 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 * \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 * \frac{2}3 * \frac{1}4</math>. If <math>n=4</math>, <math>P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5</math>.
| |
| − | | |
| − | Cross-cancelling in the fractions gives <math>P(2) = \frac{1}{2*3}</math>, <math>P(3) = </math>\frac{1}{3*4}<math>, and </math>P(4) = \frac{1}{4*5}<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>.)
| |
| − | | |
| − | </math>\frac{1}{(n+1)(n)} < \frac{1}{2010}<math>
| |
| − | | |
| − | The lowest integer that satisfies the above inequality is </math>\boxed{(A) 45}$.
| |
| − | | |
| − | == See also ==
| |
| − | {{AMC10 box|year=2010|num-b=22|num-a=24|ab=A}}
| |
| − | | |
| − | [[Category:Introductory Combinatorics Problems]] | |
