Difference between revisions of "2010 AMC 10A Problems/Problem 23"
Haihongjin (talk | contribs) m (→Solution 2) |
Haihongjin (talk | contribs) m (→Solution 2) |
||
| Line 28: | Line 28: | ||
<math>\frac{1}{(n+1)(n)} < \frac{1}{2010}</math> | <math>\frac{1}{(n+1)(n)} < \frac{1}{2010}</math> | ||
| − | The lowest integer that satisfies the above inequality is \boxed{(A) 45}. | + | The lowest integer that satisfies the above inequality is <math>\boxed{(A) 45}</math>. |
== See also == | == See also == | ||
Revision as of 20:25, 28 October 2012
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 
\frac{1}{2*3}
P(3) = 
, 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 | |
| 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 | ||
