2008 AMC 10B Problems/Problem 20
Problem
The faces of a cubical die are marked with the numbers , , , , , and . The faces of another die are marked with the numbers , , , , , and . What is the probability that the sum of the top two numbers will be , , or ?
Solution 1
One approach is to write a table of all possible outcomes, do the sums, and count good outcomes.
1 3 4 5 6 8 ------------------ 1 | 2 4 5 6 7 9 2 | 3 5 6 7 8 10 2 | 3 5 6 7 8 10 3 | 4 6 7 8 9 11 3 | 4 6 7 8 9 11 4 | 5 7 8 9 10 12
We see that out of possible outcomes, give the sum of , the sum of , and the sum of , hence the resulting probability is .
Solution 2
Each die is equally likely to roll odd or even, so the probability of an odd sum is . The possible odd sums are .
So we can find the probability of rolling or instead and just subtract that from , which seems easier.
Without writing out a table, we can see that there are 2 ways to make , and 2 ways to make , for a probability of .
.
Solution 3
The outcome of the first dice can be , , , or . For each of these cases, we can find the possible outcomes for the second dice that makes the sum of the top two numbers , , or , and then calculate the respective probabilities.
Case 1 - the first dice is 1: the outcome of the second dice can be , , or . There is a probability of rolling a with the first dice and a probability of rolling a , , or with the second dice.
Case 2 - the first dice is 2: the outcome of the second dice can be or . There is a probability of rolling a with the first dice and a probability of rolling a or with the second dice.
Case 3 - the first dice is 3: the outcome of the second dice can be or . There is a probability of rolling a with the first dice and a probability of rolling a or with the second dice.
Case 4 - the first dice is 4: the outcome of the second dice can be , , or . There is a probability of rolling a with the first dice and a probability of rolling a , , or with the second dice.
, so the answer is . ~azc1027
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.