Difference between revisions of "2006 AMC 10B Problems/Problem 21"
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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> | <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 == | ||
| − | 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 |
<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>. | <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>. | ||
Revision as of 12:46, 18 July 2006
Problem
For a particular peculiar pair of dice, the probabilities of rolling 
, 
, 
, 
, 
, and 
, on each die are in the ratio 
. What is the probability of rolling a total of 
on the two dice?
Solution
Let 
be the probability of rolling a . The probabilities of rolling a
, , , , and are 
, 
, 
, 
, and 
.
Since the sum of the probabilities of rolling each number must equal 1:
So the probabilities of rolling a , , , , , and are 
, 
,
,
,
,
.
The possible combinations of two rolls that total are: 
The probability of rolling a total of on the two dice is equal to the sum of the probabilities of rolling each combination.
