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

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==Problem==
 
==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 <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.
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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,\text{ and }8</math>. The numbers on the faces of the other die are <math>1, 2, 2, 3, 3,\text{ and }4</math>. Find the [[probability]] of rolling a sum of <math>9</math> with these two dice.
  
 
==Solution==
 
==Solution==

Latest revision as of 20:10, 5 January 2008

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 $1, 3, 4, 5, 6,\text{ and }8$. The numbers on the faces of the other die are $1, 2, 2, 3, 3,\text{ and }4$. 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
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