Difference between revisions of "2010 AMC 12A Problems/Problem 19"
Flamedragon (talk | contribs) (→Solution) |
|||
Line 4: | Line 4: | ||
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math> | <math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005</math> | ||
− | ==Solution== | + | == Solution 1== |
The probability of drawing a white marble from box <math>k</math> is <math>\frac{k}{k + 1}</math>, and the probability of drawing a red marble from box <math>k</math> is <math>\frac{1}{k+1}</math>. | The probability of drawing a white marble from box <math>k</math> is <math>\frac{k}{k + 1}</math>, and the probability of drawing a red marble from box <math>k</math> is <math>\frac{1}{k+1}</math>. | ||
Line 13: | Line 13: | ||
Since <math>n(n+1)</math> increases as <math>n</math> increases, we can simply test values of <math>n</math>; after some trial and error, we get that the minimum value of <math>n</math> is <math>\boxed{\textbf{(A) }45}</math>, since <math>45(46) = 2070</math> but <math>44(45) = 1980.</math> | Since <math>n(n+1)</math> increases as <math>n</math> increases, we can simply test values of <math>n</math>; after some trial and error, we get that the minimum value of <math>n</math> is <math>\boxed{\textbf{(A) }45}</math>, since <math>45(46) = 2070</math> but <math>44(45) = 1980.</math> | ||
+ | == Solution 2(cheap) == | ||
+ | Do the same thing as Solution 1, but when we get to <math>n(n+1)>2010</math> just test all the answer choices in ascending order(from A to E), and stop when one of the answer choices is greater than <math>2010</math>. We get <math>45(46)=2070</math>, which is greater than <math>2010</math>, so we are done. The answer is <math>\textbf{(A)}</math> | ||
+ | |||
+ | -vsamc | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}} | {{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}} |
Revision as of 11:07, 11 March 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.