Difference between revisions of "2001 AMC 8 Problems/Problem 18"
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<math>\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}</math> | <math>\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}</math> | ||
| − | ==Solution== | + | ==Solution== |
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This is equivalent to asking for the probability that at least one of the numbers is a multiple of <math> 5 </math>, since if one of the numbers is a multiple of <math> 5 </math>, then the product with it and another integer is also a multiple of <math> 5 </math>, and if a number is a multiple of <math> 5 </math>, then since <math> 5 </math> is prime, one of the factors must also have a factor of <math> 5 </math>, and <math> 5 </math> is the only multiple of <math> 5 </math> on a die, so one of the numbers rolled must be a <math> 5 </math>. To find the probability of rolling at least one <math> 5 </math>, we can find the probability of not rolling a <math> 5 </math> and subtract that from <math> 1 </math>, since you either roll a <math> 5 </math> or not roll a <math> 5 </math>. The probability of not rolling a <math> 5 </math> on either dice is <math> \left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36} </math>. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of <math> 5 </math>, is <math> 1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}} </math> | This is equivalent to asking for the probability that at least one of the numbers is a multiple of <math> 5 </math>, since if one of the numbers is a multiple of <math> 5 </math>, then the product with it and another integer is also a multiple of <math> 5 </math>, and if a number is a multiple of <math> 5 </math>, then since <math> 5 </math> is prime, one of the factors must also have a factor of <math> 5 </math>, and <math> 5 </math> is the only multiple of <math> 5 </math> on a die, so one of the numbers rolled must be a <math> 5 </math>. To find the probability of rolling at least one <math> 5 </math>, we can find the probability of not rolling a <math> 5 </math> and subtract that from <math> 1 </math>, since you either roll a <math> 5 </math> or not roll a <math> 5 </math>. The probability of not rolling a <math> 5 </math> on either dice is <math> \left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36} </math>. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of <math> 5 </math>, is <math> 1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}} </math> | ||
Revision as of 15:46, 23 December 2021
Problem
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
Solution
This is equivalent to asking for the probability that at least one of the numbers is a multiple of 
, since if one of the numbers is a multiple of , then the product with it and another integer is also a multiple of , and if a number is a multiple of , then since is prime, one of the factors must also have a factor of , and is the only multiple of on a die, so one of the numbers rolled must be a . To find the probability of rolling at least one , we can find the probability of not rolling a and subtract that from 
, since you either roll a or not roll a . The probability of not rolling a on either dice is 
. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of , is 
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
