2005 AMC 10B Problems/Problem 12
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
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 .
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 .
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)
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