2006 AMC 12B Problems/Problem 17: Difference between revisions
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== Problem == | == Problem == | ||
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? | |||
<math> | |||
\mathrm{(A)}\ \frac 4{63} | |||
\qquad | |||
\mathrm{(B)}\ \frac 18 | |||
\qquad | |||
\mathrm{(C)}\ \frac 8{63} | |||
\qquad | |||
\mathrm{(D)}\ \frac 16 | |||
\qquad | |||
\mathrm{(E)}\ \frac 27 | |||
</math> | |||
== Solution == | == Solution == | ||
The probability of getting an <math>x</math> on one of these dice is <math>\frac{x}{21}</math>. | |||
The probability of getting <math>1</math> on the first and <math>6</math> on the second die is <math>\frac 1{21}\cdot\frac 6{21}</math>. Similarly we can express the probabilities for the other five ways how we can get a total <math>7</math>. (Note that we only need the first three, the other three are symmetric.) | |||
Summing these, the probability of getting a total <math>7</math> is: | |||
<cmath> | |||
2\cdot\left( | |||
\frac 1{21}\cdot\frac 6{21} | |||
+ | |||
\frac 2{21}\cdot\frac 5{21} | |||
+ | |||
\frac 3{21}\cdot\frac 4{21} | |||
\right) | |||
= | |||
\frac{56}{441} | |||
= | |||
\boxed{\frac{8}{63}} | |||
</cmath> | |||
See also [[2016 AIME I Problems/Problem 2]] | |||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=16|num-a=18}} | |||
{{MAA Notice}} | |||
Latest revision as of 21:10, 9 June 2016
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
The probability of getting an 
on one of these dice is 
.
The probability of getting on the first and on the second die is 
. Similarly we can express the probabilities for the other five ways how we can get a total . (Note that we only need the first three, the other three are symmetric.)
Summing these, the probability of getting a total is:
![\[2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) = \frac{56}{441} = \boxed{\frac{8}{63}}\]](http://latex.artofproblemsolving.com/3/b/4/3b43693ad424a8da09fa5c6c2889616b9ea0f010.png)
See also 2016 AIME I Problems/Problem 2
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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