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 $1-\frac{25}{36}=\frac{11}{36}, \text{D}$ 
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of $5 ?$ (2001 AMC 8 Problem, Question \#14) 

a) $\frac{1}{36}$ 
b) $\frac{1}{18}$ 
c) $\frac{1}{6}$ 
d) $\frac{11}{36}$ 
e) $\frac{1}{3}$ 