Difference between revisions of "2010 AMC 12A Problems/Problem 19"
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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 | ||
| − | < | + | <center> |
| − | \frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010} | + | <math>\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}</math> |
| − | (n+1)n > 2010</ | + | |
| + | <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>. | 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>. | ||
Revision as of 14:03, 2 April 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
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 | |
