Difference between revisions of "2006 AMC 10B Problems/Problem 21"
m (added category) |
m |
||
Line 7: | Line 7: | ||
== 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>, respectively. |
− | + | 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> | ||
Line 17: | Line 17: | ||
<math>x=\frac{1}{21}</math> | <math>x=\frac{1}{21}</math> | ||
− | 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} | + | 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 ,
,
,
,
, 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
, respectively.
The sum of the probabilities of rolling each number must equal 1, so
So the probabilities of rolling a ,
,
,
,
, and
are respectively
, and
.
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.