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> | ||
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+ | == Solution 1 == | ||
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+ | 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>. |
Revision as of 15:18, 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 1
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 .