2005 AMC 10B Problems/Problem 12
Contents
[hide]Problem
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
Solution 1
In order for the product of the numbers to be prime, of the dice have to be a , and the other die has to be a prime number. There are prime numbers (, , and ), and there is only one , and there are ways to choose which die will have the prime number, so the probability is .
Solution 2
There are three cases where the product of the numbers is prime. One die will show , , or and each of the other dice will show a . For each of these three cases, the number of ways to order the numbers is = . There are possible numbers for each of the dice, so the total number of permutations is . The probability the product is prime is therefore .
~mobius247
Solution 3
The only way to get a product of that is a prime number is to roll all ones except for such prime, e.g: ones and two. So we seek the probability of rolling ones and prime number. The probability of rolling ones is and the probability of rolling a prime is , giving us a probability of of this outcome occuring. However, there are ways to arrange the ones and the prime. Multiplying the previous probability by gives us
-Benedict T (countmath1)
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AMC 10 Problems and Solutions |
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