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. Unknown error_msg)
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 | |
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All AMC 10 Problems and Solutions |