Difference between revisions of "2006 AMC 10B Problems/Problem 21"
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m (Just a clarification of the 3 cases.) |
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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> | ||
− | The probability of rolling a total of <math>7</math> on the two dice is equal to the sum of the probabilities of rolling each combination. | + | The probability P of rolling a total of <math>7</math> on the two dice is equal to the sum of the probabilities of rolling each combination. |
<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> | ||
+ | |||
+ | == Solution 2 (Not as bashy)== | ||
+ | (Alcumus solution) | ||
+ | On each die the probability of rolling <math>k</math>, for <math>1\leq | ||
+ | k\leq 6</math>, is <math> | ||
+ | \frac{k}{1+2+3+4+5+6}=\frac{k}{21}. | ||
+ | </math>There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs <math>(1,6)</math>, <math>(2,5)</math>, <math>(3,4)</math>, <math>(4,3)</math>, <math>(5,2)</math>, and <math>(6,1)</math>. Thus the probability of rolling a total of 7 is <math> | ||
+ | \frac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.</math> | ||
+ | |||
+ | ==Solution 3 (intuitive)== | ||
+ | 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> (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> (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> (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> | ||
+ | |||
+ | ~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 == | == See Also == | ||
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[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
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.