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 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. | + | [[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== | ||
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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>. | ||
| − | 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 | + | 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}} | ||
| − | + | [[Category:Intermediate Combinatorics Problems]] | |
| − | |||
Latest revision as of 00:30, 3 January 2023
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 
. The numbers on the faces of the other die are 
. Find the probability of rolling a sum of 
with these two dice.
Solution
We use generating functions to represent the sum of the two dice rolls:
The coefficient of 
, that is, the number of ways of rolling a sum of 9, is thus 
, out of a total of 
possible two-roll combinations, for a probability of 
.
Alternatively, just note the possible pairs which work: 
and 
are all possible combinations that give us a sum of (where we count 
twice because there are two different 
s to roll). Thus the probability of one of these outcomes is 
.
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 | ||
