Difference between revisions of "2006 AMC 12B Problems/Problem 17"
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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}} | {{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:
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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