2010 AMC 10A Problems/Problem 23
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
\begin{align*}\frac{1}{n+1} \left( \prod_{k&=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}\\ \frac{1}{n+1} \left( \frac{1}{n} \right) &< \frac{1}{2010}\\ (n+1)n &> 2010\end{align*} (Error compiling LaTeX. ! Missing } inserted.)
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 |