Difference between revisions of "2006 AMC 12B Problems/Problem 17"

 
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== Problem ==
 
== Problem ==
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For a particular peculiar pair of dice, the probabilities of rolling <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice?
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<math>
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\mathrm{(A)}\ \frac 4{63}
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\qquad
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\mathrm{(B)}\ \frac 18
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\qquad
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\mathrm{(C)}\ \frac 8{63}
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\qquad
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\mathrm{(D)}\ \frac 16
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\qquad
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\mathrm{(E)}\ \frac 27
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</math>
  
 
== Solution ==
 
== Solution ==
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The probability of getting an <math>x</math> on one of these dice is <math>\frac{x}{21}</math>.
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The probability of getting <math>1</math> on the first and <math>6</math> on the second die is <math>\frac 1{21}\cdot\frac 6{21}</math>. Similarly we can express the probabilities for the other five ways how we can get a total <math>7</math>. (Note that we only need the first three, the other three are symmetric.)
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Summing these, the probability of getting a total <math>7</math> is:
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<cmath>
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2\cdot\left(
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\frac 1{21}\cdot\frac 6{21}
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+
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\frac 2{21}\cdot\frac 5{21}
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+
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\frac 3{21}\cdot\frac 4{21}
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\right)
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=
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\frac{56}{441}
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=
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\boxed{\frac{8}{63}}
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</cmath>
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See also [[2016 AIME I Problems/Problem 2]]
  
 
== See also ==
 
== See also ==
* [[2006 AMC 12B Problems]]
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{{AMC12 box|year=2006|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 21:10, 9 June 2016

Problem

For a particular peculiar pair of dice, the probabilities of rolling $1$, $2$, $3$, $4$, $5$ and $6$ on each die are in the ratio $1:2:3:4:5:6$. What is the probability of rolling a total of $7$ on the two dice?

$\mathrm{(A)}\ \frac 4{63} \qquad \mathrm{(B)}\ \frac 18  \qquad \mathrm{(C)}\ \frac 8{63} \qquad \mathrm{(D)}\ \frac 16 \qquad \mathrm{(E)}\ \frac 27$

Solution

The probability of getting an $x$ on one of these dice is $\frac{x}{21}$.

The probability of getting $1$ on the first and $6$ on the second die is $\frac 1{21}\cdot\frac 6{21}$. Similarly we can express the probabilities for the other five ways how we can get a total $7$. (Note that we only need the first three, the other three are symmetric.)

Summing these, the probability of getting a total $7$ is: \[2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) =  \frac{56}{441} = \boxed{\frac{8}{63}}\]

See also 2016 AIME I Problems/Problem 2

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 12 Problems and Solutions

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