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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>
  
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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.  
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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>
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== See Also ==
 
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*[[2006 AMC 10B Problems]]
 
  
−
*[[2006 AMC 10B Problems/Problem 20|Previous Problem]]
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== Solution 2 (Not as bashy)==
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(Alcumus solution)
 +
On each die the probability of rolling <math>k</math>, for <math>1\leq
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k\leq 6</math>, is <math>
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\frac{k}{1+2+3+4+5+6}=\frac{k}{21}.
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</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>
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\frac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.</math>
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==Solution 3 (intuitive)==
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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.
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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>.
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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>.
 +
 
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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>.
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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
 +
 
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~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
 +
 
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== Video Solution by OmegaLearn ==
 +
https://youtu.be/IRyWOZQMTV8?t=3057
 +
 
 +
~ pi_is_3.14
 +
 
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== Video Solution ==
 +
https://www.youtube.com/watch?v=SBwVVADk1Nk    ~David
  
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*[[2006 AMC 10B Problems/Problem 22|Next Problem]]
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== See Also ==
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{{AMC10 box|year=2006|ab=B|num-b=20|num-a=22}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 03:05, 21 October 2024

Problem

For a particular peculiar pair of dice, the probabilities of rolling $1$, $2$, $3$, $4$, $5$, and $6$, on each die are in the ratio $1:2:3:4:5:6$. What is the probability of rolling a total of $7$ on the two dice?

$\mathrm{(A) \ } \frac{4}{63}\qquad \mathrm{(B) \ } \frac{1}{8}\qquad \mathrm{(C) \ } \frac{8}{63}\qquad \mathrm{(D) \ } \frac{1}{6}\qquad \mathrm{(E) \ } \frac{2}{7}$


Solution

Let $x$ be the probability of rolling a $1$. The probabilities of rolling a $2$, $3$, $4$, $5$, and $6$ are $2x$, $3x$, $4x$, $5x$, and $6x$, respectively.

The sum of the probabilities of rolling each number must equal 1, so

$x+2x+3x+4x+5x+6x=1$

$21x=1$

$x=\frac{1}{21}$

So the probabilities of rolling a $1$, $2$, $3$, $4$, $5$, and $6$ are respectively $\frac{1}{21}, \frac{2}{21}, \frac{3}{21}, \frac{4}{21}, \frac{5}{21}$, and $\frac{6}{21}$.

The possible combinations of two rolls that total $7$ are: $(1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)$

The probability P of rolling a total of $7$ on the two dice is equal to the sum of the probabilities of rolling each combination.

$P = \frac{1}{21}\cdot\frac{6}{21}+\frac{2}{21}\cdot\frac{5}{21}+\frac{3}{21}\cdot\frac{4}{21}+\frac{4}{21}\cdot\frac{3}{21}+\frac{5}{21}\cdot\frac{2}{21}+\frac{6}{21}\cdot\frac{1}{21}=\frac{8}{63} \Rightarrow C$

Solution 2 (Not as bashy)

(Alcumus solution) On each die the probability of rolling $k$, for $1\leq k\leq 6$, is $\frac{k}{1+2+3+4+5+6}=\frac{k}{21}.$There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, and $(6,1)$. Thus the probability of rolling a total of 7 is $\frac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.$

Solution 3 (intuitive)

There are $6$ ways to get the sum of $7$ of the dice. Let's do case by case.

Case $1$ (Rolling a $1$ or a $6$): $\frac {1}{21} \cdot \frac {6}{21} = \frac {6}{441}$.

Case $2$ (Rolling a $2$ or a $5$): $\frac {2}{21} \cdot \frac {5}{21} = \frac {10}{441}$.

Case $3$ (Rolling a $3$ or a $4$): $\frac {3}{21} \cdot \frac {4}{21} = \frac {12}{441}$.

The rest of the cases are symmetric to these cases above. We have $2 \cdot \frac {28}{441}$. We have $\frac {56}{441} = \frac {8}{63}$. Therefore, our answer is $\boxed {\frac {8}{63}}$

~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)
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Problem 20 		Followed by
Problem 22
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All AMC 10 Problems and Solutions

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