Difference between revisions of "2010 AMC 10A Problems/Problem 23"
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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) | + | <math>\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}</math> |
− | (n+1)n | + | |
+ | <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 15:03, 2 April 2010
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
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 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |