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
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 ,
, and
. From this, it is clear that
. (Alternatively,
.)
The lowest integer that satisfies the above inequality is .
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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