Difference between revisions of "2010 AMC 10A Problems/Problem 23"
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Using the first few values of <math>n</math>, it is easy to derive a formula for <math>P(n)</math>. The chance that she stops on the second box (<math>n=2</math>) is the chance of drawing a white marble then a red marble: <math>\frac{1}2 * \frac{1}3</math>. The chance that she stops on the third box (<math>n=3</math>) is the chance of drawing two white marbles then a red marble:<math>\frac{1}2 * \frac{2}3 * \frac{1}4</math>. If <math>n=4</math>, <math>P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5</math>. | Using the first few values of <math>n</math>, it is easy to derive a formula for <math>P(n)</math>. The chance that she stops on the second box (<math>n=2</math>) is the chance of drawing a white marble then a red marble: <math>\frac{1}2 * \frac{1}3</math>. The chance that she stops on the third box (<math>n=3</math>) is the chance of drawing two white marbles then a red marble:<math>\frac{1}2 * \frac{2}3 * \frac{1}4</math>. If <math>n=4</math>, <math>P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5</math>. | ||
− | Cross-cancelling in the fractions gives <math>P(2) = </math>\frac{1}{2*3}<math>, </math>P(3) = <math>\frac{1}{3*4} | + | Cross-cancelling in the fractions gives <math>P(2) = </math>\frac{1}{2*3}<math>, </math>P(3) = <math>\frac{1}{3*4}</math>, and <math>P(4) = \frac{1}{4*5}</math>. From this, it is clear that <math>P(n) = \frac{1}{(n)(n+1)}</math>. (Alternatively, <math>P(n) = \frac{(n-1)!}{(n+1)!}</math>.) |
<math>\frac{1}{(n+1)(n)} < \frac{1}{2010}</math> | <math>\frac{1}{(n+1)(n)} < \frac{1}{2010}</math> | ||
− | The lowest integer that satisfies the above inequality is 45. | + | The lowest integer that satisfies the above inequality is \boxed{(A) 45}. |
== See also == | == See also == |
Revision as of 21:25, 28 October 2012
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
Solution 1
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 .
Solution 2
Using the first few values of , it is easy to derive a formula for . The chance that she stops on the second box () is the chance of drawing a white marble then a red marble: . The chance that she stops on the third box () is the chance of drawing two white marbles then a red marble:. If , .
Cross-cancelling in the fractions gives \frac{1}{2*3}P(3) = , and . From this, it is clear that . (Alternatively, .)
The lowest integer that satisfies the above inequality is \boxed{(A) 45}.
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 |