2014 AMC 10B Problems/Problem 16
Problem
Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?
Solution
We split this problem into cases.
First, we calculate the probability that all four are the same. After the first dice, all the numbers must be equal to that roll, giving a probability of .
Second, we calculate the probability that three are the same and one is different. After the first dice, the next two must be equal and the third different. There are orders to roll the different dice, giving .
Adding these up, we get , or .
Solution 2
Note that there are two cases for this problem
: Exactly three of the dices show the same value.
There are values that the remaining die can take on, and there are ways to choose the die. There are ways that this can happen. Hence, ways.
: Exactly four of the dices show the same value.
This can happen in ways.
Hence, the probability is
Solution 3
We solve using PIE.
We first calculate the number of ways that we can have dice be the same and the other dice be anything. We therefore have ways to have at least dice be the same.
But wait! We have overcounted the case where all dice are the same! Since the previous case occurs in each of these cases times, we must subtract the -dice total three times in order to have them counted once. There are ways to have four dice be the same, so we our total count is .
Therefore, our probability is , which is answer choice .
-FIREDRAGONMATH16
Solution 4
There are two cases to consider: Three of the dice roll the same number, and all four of the dice roll the same number.
For the first case, there is a chance that one number will be rolled four times in a row. Since there are six numbers on a die, we multiply by to see that the probability for the first case is
For the second case, consider the roll , where three of the dice are identical and the fourth differs. The probability of the first three rolling the same number is because the first number can be anything, and the second must be identical. The probability of the last roll being different is , as it can be anything except for what has been previously rolled.
Multiplying these together, the probability for the second case is However, there are ways to arrange , so we must multiply by a factor of 4 to get the true probability for this case, which is
Adding these two cases, we get the requested probability: or answer choice
-Benedict T (countmath1)
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.