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

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

Latest revision as of 18:53, 31 October 2016

Problem

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 (and equally likely) values Tamika gets are:

$8+9=17$

$8+10=18$

$9+10=19$

The different possible (and equally likely) values Carlos gets are:

$3\times5=15$

$3\times6=18$

$5\times6=30$

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

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