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Difference between revisions of "1992 AJHSME Problems/Problem 23"

(Created page with "==Problem== If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than 10 is <math> \text{(A)}\ \frac{3}{7}\qquad\t...")
 
 
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<math> \text{(A)}\ \frac{3}{7}\qquad\text{(B)}\ \frac{17}{36}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{12} </math>
 
<math> \text{(A)}\ \frac{3}{7}\qquad\text{(B)}\ \frac{17}{36}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{12} </math>
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==Solution==
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When the first dice is <math>1</math>, there are no possibilities for the second dice.
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When the first dice is <math>2</math>, the second dice can be <math>6</math>, giving one possibility.
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When the first dice is <math>3</math>, the second dice can be <math>4,5,6</math>, giving three possibilities.
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When the first dice is <math>4</math>, the second dice can be <math>3,4,5,6</math>, giving four possibilities.
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When the first dice is <math>5</math>, the second dice can be <math>3,4,5,6</math>, giving four possibilities.
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When the first dice is <math>6</math>, the second dice can be <math>2,3,4,5,6</math>, giving five possibilities.
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The total number of ways for the product to greater than 10 is <math>1+3+4+4+5=17</math> and the total number of possibilities is <math>(6)(6)=36</math>, yielding a probability of <math>\boxed{\text{(B)}\ \frac{17}{36}}</math>.
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==See Also==
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{{AJHSME box|year=1992|num-b=22|num-a=24}}
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{{MAA Notice}}

Latest revision as of 00:10, 5 July 2013

Problem

If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than 10 is

$\text{(A)}\ \frac{3}{7}\qquad\text{(B)}\ \frac{17}{36}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{12}$


Solution

When the first dice is $1$, there are no possibilities for the second dice.

When the first dice is $2$, the second dice can be $6$, giving one possibility.

When the first dice is $3$, the second dice can be $4,5,6$, giving three possibilities.

When the first dice is $4$, the second dice can be $3,4,5,6$, giving four possibilities.

When the first dice is $5$, the second dice can be $3,4,5,6$, giving four possibilities.

When the first dice is $6$, the second dice can be $2,3,4,5,6$, giving five possibilities.


The total number of ways for the product to greater than 10 is $1+3+4+4+5=17$ and the total number of possibilities is $(6)(6)=36$, yielding a probability of $\boxed{\text{(B)}\ \frac{17}{36}}$.

See Also

1992 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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