Difference between revisions of "1992 AJHSME Problems/Problem 23"
Mrdavid445 (talk | contribs) (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== | ||
| + | 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== | ||
| + | {{AJHSME box|year=1992|num-b=22|num-a=24}} | ||
Revision as of 21:11, 22 December 2012
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
Solution
When the first dice is 
, there are no possibilities for the second dice.
When the first dice is 
, the second dice can be 
, giving one possibility.
When the first dice is 
, the second dice can be 
, giving three possibilities.
When the first dice is 
, the second dice can be 
, giving four possibilities.
When the first dice is 
, the second dice can be , giving four possibilities.
When the first dice is , the second dice can be 
, giving five possibilities.
The total number of ways for the product to greater than 10 is 
and the total number of possibilities is 
, yielding a probability of 
.
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 | ||
