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Difference between revisions of "2005 Alabama ARML TST Problems/Problem 1"

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==Problem==
 
==Problem==
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Two six-sided dice are constructed such that each face is equally likely to show up when rolled. The numbers on the faces of one of the dice are 1, 3, 4, 5, 6, and 8. The numbers on the faces of the other die are 1, 2, 2, 3, 3, and 4. Find the [[probability]] of rolling a sum of 9 with these two dice.
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Two six-sided dice are constructed such that each face is equally likely to show up when rolled. The numbers on the faces of one of the dice are <math>1, 3, 4, 5, 6, and 8</math>. The numbers on the faces of the other die are <math>1, 2, 2, 3, 3, and 4</math>. Find the [[probability]] of rolling a sum of <math>9</math> with these two dice.
  
 
==Solution==
 
==Solution==
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The [[coefficient]] of <math>x^9</math>, that is, the number of ways of rolling a sum of 9, is thus <math>(1+2+1)=4</math>, out of a total of <math>6^2</math> possible two-roll combinations, for a probability of <math>\frac 19</math>.
 
The [[coefficient]] of <math>x^9</math>, that is, the number of ways of rolling a sum of 9, is thus <math>(1+2+1)=4</math>, out of a total of <math>6^2</math> possible two-roll combinations, for a probability of <math>\frac 19</math>.
  
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Alternatively, just note the possible pairs which work: <math>(5, 4), (6, 3), (6, 3)</math> and <math>(8, 1)</math> are all possible combinations that give us a sum of 9 (where we count <math>(6, 3)</math> twice because there are two different 3s to roll).  Thus the probability of one of these outcomes is <math>\frac{4}{36} = \frac19</math>.
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Alternatively, just note the possible pairs which work: <math>(5, 4), (6, 3), (6, 3)</math> and <math>(8, 1)</math> are all possible combinations that give us a sum of <math>9</math> (where we count <math>(6, 3)</math> twice because there are two different <math>3</math>s to roll).  Thus the probability of one of these outcomes is <math>\frac{4}{36} = \frac19</math>.
  
 
==See also==
 
==See also==
 
{{ARML box|year=2005|state=Alabama|before=First question|num-a=2}}
 
{{ARML box|year=2005|state=Alabama|before=First question|num-a=2}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 11:41, 11 December 2007

Problem

Two six-sided dice are constructed such that each face is equally likely to show up when rolled. The numbers on the faces of one of the dice are $1, 3, 4, 5, 6, and 8$. The numbers on the faces of the other die are $1, 2, 2, 3, 3, and 4$. Find the probability of rolling a sum of $9$ with these two dice.

Solution

We use generating functions to represent the sum of the two dice rolls:

$(x+x^3+x^4+x^5+x^6+x^8)(x+2x^2+2x^3+x^4)=$
$x^2(1+x^2+x^3+x^4+x^5+x^7)(1+x+x^2)(1+x)$

The coefficient of $x^9$, that is, the number of ways of rolling a sum of 9, is thus $(1+2+1)=4$, out of a total of $6^2$ possible two-roll combinations, for a probability of $\frac 19$.

Alternatively, just note the possible pairs which work: $(5, 4), (6, 3), (6, 3)$ and $(8, 1)$ are all possible combinations that give us a sum of $9$ (where we count $(6, 3)$ twice because there are two different $3$s to roll). Thus the probability of one of these outcomes is $\frac{4}{36} = \frac19$.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
First question 		Followed by:
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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