Difference between revisions of "2018 AMC 10B Problems/Problem 6"
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<math>\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}</math> | <math>\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}</math> | ||
| − | == Solution | + | == Solution == |
Notice that the only two ways such that more than <math>2</math> draws are required are <math>1,2</math>, <math>1,3</math>, <math>2,1</math>, and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>. | Notice that the only two ways such that more than <math>2</math> draws are required are <math>1,2</math>, <math>1,3</math>, <math>2,1</math>, and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2018|ab=B|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:28, 16 February 2018
A box contains 
chips, numbered 
, 
, 
, 
, and . Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds . What is the probability that draws are required?
Solution
Notice that the only two ways such that more than draws are required are 
, 
, 
, and 
. Notice that each of those cases has a 
chance, so the answer is 
, or 
.
See Also
| 2018 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
