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

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== Solution ==
 
== Solution ==
 
Let <math>x</math> be the [[probability]] of rolling a <math>1</math>. The probabilities of rolling a  
 
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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<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>, respectively.
  
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Since the sum of the probabilities of rolling each number must equal 1:
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The sum of the probabilities of rolling each number must equal 1, so
  
 
<math>x+2x+3x+4x+5x+6x=1</math>
 
<math>x+2x+3x+4x+5x+6x=1</math>
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<math>x=\frac{1}{21}</math>
 
<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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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 respectively <math>\frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}</math>, and <math>\frac{6}{21}</math>.  
  
 
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>
 
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>

Revision as of 13:56, 14 August 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 $6x$, respectively.

The sum of the probabilities of rolling each number must equal 1, so

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

$21x=1$

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

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

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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