Difference between revisions of "2005 AMC 10B Problems/Problem 12"
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+ | == Solution 3== | ||
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+ | The only way to get a product of that is a prime number is to roll all ones except for such prime, e.g: <math>11</math> ones and <math>1</math> two. So we seek the probability of rolling <math>11</math> ones and <math>1</math> prime number. The probability of rolling <math>11</math> ones is <math>\frac{1}{6^{11}}</math> and the probability of rolling a prime is <math>\frac{1}{2}</math>, giving us a probability of <math>\frac{1}{6^{11}}\cdot\frac{1}{2}</math> of this outcome occuring. However, there are <math>\frac{12!}{11!\cdot{1!}}=12</math> ways to arrange the ones and the prime. Multiplying the previous probability by <math>12</math> gives us <math>\frac{1}{6^{11}}\cdot\frac{1}{2}\cdot{6}=\frac{1}{6^{10}}=\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}.</math> | ||
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+ | -Benedict T (countmath1) | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2005|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:03, 16 June 2022
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 | |
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 AMC 10 Problems and Solutions |
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