2010 AMC 12A Problems/Problem 19
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 1
The probability of drawing a white marble from box is
, and the probability of drawing a red marble from box
is
.
To stop after drawing marbles, we must draw a white marble from boxes
and draw a red marble from box
Thus,
So, we must have or
Since increases as
increases, we can simply test values of
; after some trial and error, we get that the minimum value of
is
, since
but
Solution 2(cheap)
Do the same thing as Solution 1, but when we get to just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than
. We get
, which is greater than
, so we are done. The answer is
-vsamc
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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