Difference between revisions of "2008 AMC 10B Problems/Problem 20"

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==Problem==
 
==Problem==
The faces of a cubical die are marked with the numbers 1, 2, 2, 3, 3, and 4. The faces of another die are marked with the numbers 1, 3, 4, 5, 6, and 8. What is the probability that the sum of the top two numbers will be 5, 7, or 9?                   
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The faces of a cubical die are marked with the numbers <math>1</math>, <math>2</math>, <math>2</math>, <math>3</math>, <math>3</math>, and <math>4</math>. The faces of another die are marked with the numbers <math>1</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, and <math>8</math>. What is the probability that the sum of the top two numbers will be <math>5</math>, <math>7</math>, or <math>9</math>?                   
(A) 5/18 (B) 7/18 (C) 11/18 (D) 3/4 (E) 8/9  
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<math>\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9</math>
  
 
==Solution==
 
==Solution==

Revision as of 16:05, 25 January 2009

Problem

The faces of a cubical die are marked with the numbers $1$, $2$, $2$, $3$, $3$, and $4$. The faces of another die are marked with the numbers $1$, $3$, $4$, $5$, $6$, and $8$. What is the probability that the sum of the top two numbers will be $5$, $7$, or $9$?

$\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9$

Solution

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See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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