Difference between revisions of "2005 AMC 10B Problems/Problem 12"
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Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime? | Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime? | ||
− | <math>\ | + | <math>\textbf{(A) } \left(\frac{1}{12}\right)^{12} \qquad \textbf{(B) } \left(\frac{1}{6}\right)^{12} \qquad \textbf{(C) } 2\left(\frac{1}{6}\right)^{11} \qquad \textbf{(D) } \frac{5}{2}\left(\frac{1}{6}\right)^{11} \qquad \textbf{(E) } \left(\frac{1}{6}\right)^{10} </math> |
− | == Solution == | + | |
− | In order for the product of the numbers to be prime, <math>11</math> of the dice have to be a <math>1</math>, and the other die has to be a prime number. There are <math>3</math> prime numbers (<math>2</math>, <math>3</math>, and <math>5</math>), and there is only one <math>1</math>, and there are <math>\dbinom{12}{1}</math> ways to choose which die will have the prime number, so the probability is <math>\dfrac{3}{6}\times\left(\dfrac{1}{6}\right)^{11}\times\dbinom{12}{1} = \dfrac{1}{2}\times\left(\dfrac{1}{6}\right)^{11}\times12=\left(\dfrac{1}{6}\right)^{11}\times6=\boxed{\ | + | == Solution 1 == |
+ | In order for the product of the numbers to be prime, <math>11</math> of the dice have to be a <math>1</math>, and the other die has to be a prime number. There are <math>3</math> prime numbers (<math>2</math>, <math>3</math>, and <math>5</math>), and there is only one <math>1</math>, and there are <math>\dbinom{12}{1}</math> ways to choose which die will have the prime number, so the probability is <math>\dfrac{3}{6}\times\left(\dfrac{1}{6}\right)^{11}\times\dbinom{12}{1} = \dfrac{1}{2}\times\left(\dfrac{1}{6}\right)^{11}\times12=\left(\dfrac{1}{6}\right)^{11}\times6=\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}</math>. | ||
== Solution 2== | == Solution 2== | ||
− | There are three cases where the product of the numbers is prime. One die will show <math>2</math>, <math>3</math>, or <math>5</math> and each of the other <math>11</math> dice will show a <math>1</math>. For each of these three cases, the number of ways to order the numbers is <math>\dbinom{12}{1}</math> = <math>12</math> . There are <math>6</math> possible numbers for each of the <math>12</math> dice, so the total number of permutations is <math>6^{12}</math>. The probability the product is prime is therefore \frac{3\cdot 12}{6^{12} | + | There are three cases where the product of the numbers is prime. One die will show <math>2</math>, <math>3</math>, or <math>5</math> and each of the other <math>11</math> dice will show a <math>1</math>. For each of these three cases, the number of ways to order the numbers is <math>\dbinom{12}{1}</math> = <math>12</math> . There are <math>6</math> possible numbers for each of the <math>12</math> dice, so the total number of permutations is <math>6^{12}</math>. The probability the product is prime is therefore |
+ | <math>\frac{3\cdot 12}{6^{12}} = \frac{1}{6^{10}} =\boxed{\textbf{(E) }\left(\dfrac{1}{6}\right)^{10}}</math>. | ||
+ | |||
+ | ~mobius247 | ||
− | + | == Solution 3== | |
− | <math>\frac{ | + | 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> |
− | + | -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 14:03, 16 June 2022
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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