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Difference between revisions of "2006 AMC 10B Problems/Problem 21"

 
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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> \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} </math>
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== Solution ==
 
== Solution ==
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Let <math>x</math> be the probability of rolling a <math>1</math>. The probabilities of rolling a
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<math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are <math>2x</math>, <math>3x</math>, <math>4x</math>, <math>5x</math>, and <math>6x</math>
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Since the sum of the probabilities of rolling each number must equal 1:
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<math>x+2x+3x+4x+5x+6x=1</math>
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<math>21x=1</math>
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<math>x=\frac{1}{21}</math>
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So the probabilities of rolling a <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are <math>\frac{1}{21}</math>, <math>\frac{2}{21}</math>,<math>\frac{3}{21}</math>,<math>\frac{4}{21}</math>,<math>\frac{5}{21}</math>,<math>\frac{6}{21}</math>.
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The possible combinations of two rolls that total <math>7</math> are: <math> (1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1) </math>
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The probability of rolling a total of <math>7</math> on the two dice is equal to the sum of the probabilities of rolling each combination.
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<math> P = \frac{1}{21}\cdot\frac{6}{21}+\frac{2}{21}\cdot\frac{5}{21}+\frac{3}{21}\cdot\frac{4}{21}+\frac{4}{21}\cdot\frac{3}{21}+\frac{5}{21}\cdot\frac{2}{21}+\frac{6}{21}\cdot\frac{1}{21}=\frac{8}{63} \Rightarrow C </math>
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
*[[2006 AMC 10B Problems]]

Revision as of 12:45, 14 July 2006

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?

$\mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7}$

Solution

Let $x$ be the probability of rolling a $1$. The probabilities of rolling a $2$, $3$, $4$, $5$, and $6$ are $2x$, $3x$, $4x$, $5x$, and $6x$

Since the sum of the probabilities of rolling each number must equal 1:

$x+2x+3x+4x+5x+6x=1$

$21x=1$

$x=\frac{1}{21}$

So the probabilities of rolling a $1$, $2$, $3$, $4$, $5$, and $6$ are $\frac{1}{21}$, $\frac{2}{21}$,$\frac{3}{21}$,$\frac{4}{21}$,$\frac{5}{21}$,$\frac{6}{21}$.

The possible combinations of two rolls that total $7$ are: $(1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)$

The probability of rolling a total of $7$ on the two dice is equal to the sum of the probabilities of rolling each combination.

$P = \frac{1}{21}\cdot\frac{6}{21}+\frac{2}{21}\cdot\frac{5}{21}+\frac{3}{21}\cdot\frac{4}{21}+\frac{4}{21}\cdot\frac{3}{21}+\frac{5}{21}\cdot\frac{2}{21}+\frac{6}{21}\cdot\frac{1}{21}=\frac{8}{63} \Rightarrow C$

See Also

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