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 (countmath2)
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.
What is the probability of rolling the same number 3 times on a dice?
The three dice are rolled fairly without any cheating. Each of the dice rolls is an Independent Event, that is the outcome from anyone dice roll has no impact whatsoever on the outcome of any other dice roll. The probability of all three happening is the product of the three probabilities: 1 × (1/6) × (1/6) = 1/36.
What is the probability of rolling a 3 if you rolled a 6 sided dice?
Originally Answered: If you roll a six sided dice, what is the probability of getting 3? There are six numbers possible. The probability of getting the number 3 out of 1,2,3,4,5,6 is 1/6.
What is the probability of rolling a 5 on a 6 sided dice?
There is one side marked 5 on a standard six sided die. Assuming we are talking such a die, and that the die is honest, then the odds are 1/6. However, there's a die which is marked all 5s, useful in allowing you to always roll a 7 or 11 when combined with a second special die marked with half 2s and half 6s.