Difference between revisions of "2001 AMC 8 Problems/Problem 18"
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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> | ||
− | ==Solution 2== | + | ==Solution 2 (quick & easy)== |
− | The only way to get a multiple of 5 is to have at least one 5. If the first dice rolls a 5, there are 6 ways to get a multiple of 5. If the second dice rolls a 5, there are also 6 ways. However, | + | The only way to get a multiple of <math> 5 </math> is to have at least one <math> 5 </math>. If the first dice rolls a <math> 5 </math>, there are <math> 6 </math> ways to get a multiple of <math> 5 </math>. If the second dice rolls a <math> 5 </math>, there are also <math> 6 </math> ways. However, one case is repeated: both dice roll a <math> 5 </math>. Therefore, there are <math> 6 + 6 - 1 = 11 </math>, and there is a total of <math> 6 \times 6 </math> ways, so the probability is <math>\text{(D)}\ \dfrac{11}{36}</math> |
− | + | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] | |
==Video Solution== | ==Video Solution== |
Latest revision as of 17:24, 15 August 2024
Problem
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
Solution 1
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
Solution 2 (quick & easy)
The only way to get a multiple of is to have at least one
. If the first dice rolls a
, there are
ways to get a multiple of
. If the second dice rolls a
, there are also
ways. However, one case is repeated: both dice roll a
. Therefore, there are
, and there is a total of
ways, so the probability is
Solution by ILoveMath31415926535
Video Solution
https://youtu.be/4aX9-DZHgNw Soo, DRMS, NM
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 |
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