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 == | ||
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>? | 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>? | ||
Revision as of 19:22, 26 May 2020
- 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 
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 1
The probability of drawing a white marble from box is 
, and the probability of drawing a red marble from box is 
.
To stop after drawing marbles, we must draw a white marble from boxes 
and draw a red marble from box 
Thus, ![\[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\]](http://latex.artofproblemsolving.com/4/a/f/4afea34f3ad6f2274778795778b2d5dad8ab0e85.png)
So, we must have 
or 
Since 
increases as increases, we can simply test values of ; after some trial and error, we get that the minimum value of is 
, since 
but 
Solution 2(cheap)
Do the same thing as Solution 1, but when we get to 
just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than . We get 
, which is greater than , so we are done. The answer is 
-vsamc
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
