Difference between revisions of "2018 AMC 10B Problems/Problem 6"

(Created page with "A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips we drawn randomly one at a time without...")
 
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A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips we drawn randomly one at a time without replacement until the sun of the values drawn exceeds <math>4</math>. What is the probability that <math>3</math> draws are required?
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A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds <math>4</math>. What is the probability that <math>3</math> draws are required?
  
 
<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>

Revision as of 14:10, 16 February 2018

A box contains $5$ chips, numbered $1$, $2$, $3$, $4$, and $5$. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4$. What is the probability that $3$ draws are required?

$\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}$