Difference between revisions of "2010 AMC 12A Problems/Problem 19"
(Created page with '== Problem 19 == 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 <…') |
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| − | <math>\frac{1}{n+1}(\prod_{k=1}^{n-1}\frac{k}{k+1}) < \frac{1}{2010}</math> | + | <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}(\frac{1}{n}) < \frac{1}{2010}</math> | + | <math>\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}</math> |
Revision as of 20:47, 10 February 2010
Problem 19
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
It is then easy to see that the lowest integer value of that satisfies the inequality is 
.
