Difference between revisions of "2014 AMC 10B Problems/Problem 16"
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==Solution== | ==Solution== | ||
| − | We split this problem into 2 cases. | + | We split this problem into <math>2</math> cases. |
First, we calculate the probability that all four are the same. After the first dice, all the number must be equal to that roll, giving a probability of <math>1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{216}</math>. | First, we calculate the probability that all four are the same. After the first dice, all the number must be equal to that roll, giving a probability of <math>1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{216}</math>. | ||
| − | 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 4 orders to roll the different dice, giving <math>4 \cdot 1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{5}{6} = \dfrac{5}{54}</math>. | + | 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 <math>4</math> orders to roll the different dice, giving <math>4 \cdot 1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{5}{6} = \dfrac{5}{54}</math>. |
Adding these up, we get <math>\dfrac{7}{72}</math>, or <math>\boxed{\textbf{(B)}}</math>. | Adding these up, we get <math>\dfrac{7}{72}</math>, or <math>\boxed{\textbf{(B)}}</math>. | ||
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Note that there are two cases for this problem | Note that there are two cases for this problem | ||
| − | Case 1: Exactly three of the dices show the same value. | + | <math>\textbf{Case 1}</math>: Exactly three of the dices show the same value. |
There are <math>5</math> values that the remaining die can take on, and there are <math>\binom{4}{3}=4</math> ways to choose the die. There are <math>6</math> ways that this can happen. Hence, <math>6\cdot 4\cdot5=120</math> ways. | There are <math>5</math> values that the remaining die can take on, and there are <math>\binom{4}{3}=4</math> ways to choose the die. There are <math>6</math> ways that this can happen. Hence, <math>6\cdot 4\cdot5=120</math> ways. | ||
| − | Case 2: Exactly four of the dices show the same value. | + | <math>\textbf{Case 2}</math>: Exactly four of the dices show the same value. |
This can happen in <math>6</math> ways. | This can happen in <math>6</math> ways. | ||
Revision as of 18:52, 27 December 2019
Contents
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 number 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 
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. 
