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==
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==Solution== Complementary Counting
  
 
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 16:46, 23 December 2021

Problem

Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?

$\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}$

==Solution== Complementary Counting

This is equivalent to asking for the probability that at least one of the numbers is a multiple of $5$, since if one of the numbers is a multiple of $5$, then the product with it and another integer is also a multiple of $5$, and if a number is a multiple of $5$, then since $5$ is prime, one of the factors must also have a factor of $5$, and $5$ is the only multiple of $5$ on a die, so one of the numbers rolled must be a $5$. To find the probability of rolling at least one $5$, we can find the probability of not rolling a $5$ and subtract that from $1$, since you either roll a $5$ or not roll a $5$. The probability of not rolling a $5$ on either dice is $\left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36}$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $5$, is $1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}}$

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
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

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