Difference between revisions of "2010 AMC 12A Problems/Problem 19"
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− | It is then easy to see that the lowest integer value of <math>n</math> that satisfies the inequality is <math>\boxed{45\ \textbf{( | + | 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 23:21, 11 February 2010
Problem 19
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
.