Difference between revisions of "2006 AMC 10B Problems/Problem 21"

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(Solution 2)
 
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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>
== See Also ==
 
*[[2006 AMC 10B Problems]]
 
  
*[[2006 AMC 10B Problems/Problem 20|Previous Problem]]
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== Solution 2 ==
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(Alcumus solution)
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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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== Video Solution ==
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https://youtu.be/IRyWOZQMTV8?t=3057
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~ pi_is_3.14
  
*[[2006 AMC 10B Problems/Problem 22|Next Problem]]
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== See Also ==
 +
{{AMC10 box|year=2006|ab=B|num-b=20|num-a=22}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 20:57, 17 January 2021

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 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

(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 $\frac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.$

Video Solution

https://youtu.be/IRyWOZQMTV8?t=3057

~ pi_is_3.14

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 10 Problems and Solutions

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