Difference between revisions of "2006 AMC 10B Problems/Problem 21"
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<math> \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} </math> | <math> \mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7} </math> | ||
+ | |||
== Solution == | == Solution == | ||
− | Let <math>x</math> be the probability of rolling a <math>1</math>. The probabilities of rolling a | + | Let <math>x</math> be the [[probability]] of rolling a <math>1</math>. The probabilities of rolling a |
− | <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are <math>2x</math>, <math>3x</math>, <math>4x</math>, <math>5x</math>, and <math>6x</math> | + | <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are <math>2x</math>, <math>3x</math>, <math>4x</math>, <math>5x</math>, and <math>6x</math>, respectively. |
− | + | The sum of the probabilities of rolling each number must equal 1, so | |
<math>x+2x+3x+4x+5x+6x=1</math> | <math>x+2x+3x+4x+5x+6x=1</math> | ||
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<math>x=\frac{1}{21}</math> | <math>x=\frac{1}{21}</math> | ||
− | So the probabilities of rolling a <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are <math>\frac{1}{21} | + | So the probabilities of rolling a <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are respectively <math>\frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}</math>, and <math>\frac{6}{21}</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 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> | ||
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<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 == | ||
+ | (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> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/IRyWOZQMTV8?t=3057 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2006|ab=B|num-b=20|num-a=22}} | |
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 21:57, 17 January 2021
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.
Solution 2
(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
Video Solution
https://youtu.be/IRyWOZQMTV8?t=3057
~ pi_is_3.14
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 |
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