Difference between revisions of "2006 AMC 10B Problems/Problem 21"
m |
Yankeesfan (talk | contribs) (→Solution) |
||
| Line 24: | Line 24: | ||
<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> | <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 11:59, 17 February 2010
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.
