Difference between revisions of "2010 AMC 12A Problems/Problem 19"
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| − | == Problem | + | == Problem == |
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>? | 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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The probability of drawing a red marble at box <math>n</math> is therefore | The probability of drawing a red marble at box <math>n</math> is therefore | ||
| + | <cmath>\begin{align*}\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\end{align*}</cmath> | ||
| − | <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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| − | <math>\ | ||
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| − | + | == See also == | |
| + | {{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}} | ||
| − | + | [[Category:Introductory Combinatorics Problems]] | |
| − | |||
Revision as of 23:32, 25 February 2010
Problem
Each of 2010 boxes in a line contains a single red marble, and for 
, the box in the 
position also contains 
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 
be the probability that Isabella stops after drawing exactly 
marbles. What is the smallest value of for which 
?
Solution
The probability of drawing a white marble from box is 
. The probability of drawing a red marble from box is 
.
The probability of drawing a red marble at box is therefore
\begin{align*}\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\end{align*} (Error compiling LaTeX. Unknown error_msg)
It is then easy to see that the lowest integer value of that satisfies the inequality is 
.
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 | |
