Difference between revisions of "2006 AMC 10B Problems/Problem 21"
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There are <math>6</math> ways to get the sum of <math>7</math> of the dice. Let's do case by case. | There are <math>6</math> ways to get the sum of <math>7</math> of the dice. Let's do case by case. | ||
| − | Case <math>1</math>: <math>\frac {1}{21} \cdot \frac {6}{21} = \frac {6}{441}</math>. | + | Case <math>1</math> (Rolling a <math>1</math> or a <math>6</math>): <math>\frac {1}{21} \cdot \frac {6}{21} = \frac {6}{441}</math>. |
| − | Case <math>2</math>: <math>\frac {2}{21} \cdot \frac {5}{21} = \frac {10}{441}</math>. | + | Case <math>2</math> (Rolling a <math>2</math> or a <math>5</math>): <math>\frac {2}{21} \cdot \frac {5}{21} = \frac {10}{441}</math>. |
| − | Case <math>3</math>: <math>\frac {3}{21} \cdot \frac {4}{21} = \frac {12}{441}</math>. | + | Case <math>3</math> (Rolling a <math>3</math> or a <math>4</math>): <math>\frac {3}{21} \cdot \frac {4}{21} = \frac {12}{441}</math>. |
The rest of the cases are symmetric to these cases above. We have <math>2 \cdot \frac {28}{441}</math>. We have <math>\frac {56}{441} = \frac {8}{63}</math>. Therefore, our answer is <math>\boxed {\frac {8}{63}}</math> | The rest of the cases are symmetric to these cases above. We have <math>2 \cdot \frac {28}{441}</math>. We have <math>\frac {56}{441} = \frac {8}{63}</math>. Therefore, our answer is <math>\boxed {\frac {8}{63}}</math> | ||
~Arcticturn | ~Arcticturn | ||
| + | |||
| + | ~Minor Edits by L314159265358979323846264 | ||
==Solution 4 (cheap)== | ==Solution 4 (cheap)== | ||
Latest revision as of 03:05, 21 October 2024
Contents
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 P of rolling a total of on the two dice is equal to the sum of the probabilities of rolling each combination.
Solution 2 (Not as bashy)
(Alcumus solution)
On each die the probability of rolling 
, for 
, is 
There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs 
, 
, 
, 
, 
, and 
. Thus the probability of rolling a total of 7 is 
Solution 3 (intuitive)
There are ways to get the sum of of the dice. Let's do case by case.
Case (Rolling a or a ): 
.
Case (Rolling a or a ): 
.
Case (Rolling a or a ): 
.
The rest of the cases are symmetric to these cases above. We have 
. We have 
. Therefore, our answer is 
~Arcticturn
~Minor Edits by L314159265358979323846264
Solution 4 (cheap)
Notice that the ways to obtain a 7 are (6,1), (5,2), (4,3). Then, because the cases are symmetrical with (3,4), (2,5) and (6,1), look through the answer choices. You see a 4/63 and an 8/63, so obviously MAA wants you to not count the other symmetrical cases, thus giving the answer (C).
~mathboy282
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=3057
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=SBwVVADk1Nk ~David
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
