Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1998 AJHSME Problems/Problem 19"

(Created page, added problem, solution, navigational box)
 
m (→‎Solution: Added spaces)
Line 10: Line 10:
  
 
<math>8+9=17</math>
 
<math>8+9=17</math>
 +
 
<math>8+10=18</math>
 
<math>8+10=18</math>
 +
 
<math>9+10=19</math>
 
<math>9+10=19</math>
  
Line 16: Line 18:
  
 
<math>3\times5=15</math>
 
<math>3\times5=15</math>
 +
 
<math>3\times6=18</math>
 
<math>3\times6=18</math>
 +
 
<math>5\times6=30</math>
 
<math>5\times6=30</math>
  
Line 26: Line 30:
  
 
<math>\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{A}</math>
 
<math>\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{A}</math>
−
 
  
 
== See also ==
 
== See also ==

Revision as of 12:07, 10 June 2011

Problem 19

Tamika selects two different numbers at random from the set $\{ 8,9,10 \}$ and adds them. Carlos takes two different numbers at random from the set $\{3, 5, 6\}$ and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?

$\text{(A)}\ \dfrac{4}{9} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{1}{3} \qquad \text{(E)}\ \dfrac{2}{3}$

Solution

The different possible values of Tamika's set leaves:

$8+9=17$

$8+10=18$

$9+10=19$

The different possible values of Carlos's set leaves:

$3\times5=15$

$3\times6=18$

$5\times6=30$

The probability that if Tamika had the sum $17$ her sum would be greater than Carlos's set is $\frac{1}{3}$, because $17$ is only greater than $15$ The probability that if Tamika had the sum $18$ her sum would be greater than Carlos's set is $\frac{1}{3}$, because $18$ is only greater than $15$ The probability that if Tamika had the sum $19$ her sum would be greater than Carlos's set is $\frac{2}{3}$, because $19$ is greater than both $15$ and $18$

Each sum has a $\frac{1}{3}$ possibility of being chosen, so we have

$\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{A}$

See also

1998 AJHSME (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 AJHSME/AMC 8 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_19&oldid=39462"