Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 10B Problems/Problem 16"

m (→‎Solution)
m (minor typpo)
 
(4 intermediate revisions by 3 users not shown)
Line 4: Line 4:
 
<math> \textbf {(A) } \frac{1}{36} \qquad \textbf {(B) } \frac{7}{72} \qquad \textbf {(C) } \frac{1}{9} \qquad \textbf {(D) } \frac{5}{36} \qquad \textbf {(E) } \frac{1}{6}</math>
 
<math> \textbf {(A) } \frac{1}{36} \qquad \textbf {(B) } \frac{7}{72} \qquad \textbf {(C) } \frac{1}{9} \qquad \textbf {(D) } \frac{5}{36} \qquad \textbf {(E) } \frac{1}{6}</math>
  
−
==Solution==
+
==Solution 1==
  
 
We split this problem into <math>2</math> cases.
 
We split this problem into <math>2</math> cases.
Line 30: Line 30:
 
==Solution 3==
 
==Solution 3==
  
−
We solve using PIE.
+
We solve using [[PIE]].
  
 
We first calculate the number of ways that we can have <math>3</math> dice be the same and the other dice be anything. We therefore have <math>\binom{4}{3} \cdot 6 \cdot 6 = 144</math> ways to have at least <math>3</math> dice be the same.
 
We first calculate the number of ways that we can have <math>3</math> dice be the same and the other dice be anything. We therefore have <math>\binom{4}{3} \cdot 6 \cdot 6 = 144</math> ways to have at least <math>3</math> dice be the same.
Line 39: Line 39:
  
 
-FIREDRAGONMATH16
 
-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 <math>\frac{1}{6^4}</math> chance that one number will be rolled four times in a row. Since there are six numbers on a die, we multiply by <math>6</math> to see that the probability for the first case is <math>\frac{1}{6^3}.</math>
 +
 +
 +
For the second case, consider the roll <math>AAAB</math>, where three of the dice are identical and the fourth differs. The probability of the first three rolling the same number is <math>1\cdot{\frac{1}{6}}\cdot{\frac{1}{6}},</math> because the first number can be anything, and the second must be identical. The probability of the last roll being different is <math>\frac{5}{6}</math>, as it can be anything except for what has been previously rolled.
 +
 +
 +
Multiplying these together, the probability for the second case is <math>\frac{5}{6^3}.</math> However, there are <math>\frac{4!}{3!\cdot{1!}} = 4</math> ways to arrange <math>AAAB</math>, so we must multiply by a factor of 4 to get the true probability for this case, which is <math>4(\frac{5}{6^3}) = \frac{20}{6^3}.</math>
 +
 +
 +
Adding these two cases, we get the requested probability: <math>\frac{1+20}{6^3} = \frac{21}{216} = \frac{7}{72},</math> or answer choice <math>\boxed{\textbf{(B)}}</math>
 +
 +
 +
-Benedict T (countmath1)
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2014|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:59, 10 November 2024

Problem

Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?

$\textbf {(A) } \frac{1}{36} \qquad \textbf {(B) } \frac{7}{72} \qquad \textbf {(C) } \frac{1}{9} \qquad \textbf {(D) } \frac{5}{36} \qquad \textbf {(E) } \frac{1}{6}$

Solution 1

We split this problem into $2$ 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 $1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{216}$.

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 $4 \cdot 1 \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} \cdot \dfrac{5}{6} = \dfrac{5}{54}$.

Adding these up, we get $\dfrac{7}{72}$, or $\boxed{\textbf{(B)}}$.

Solution 2

Note that there are two cases for this problem

$\textbf{Case 1}$: Exactly three of the dices show the same value.

There are $5$ values that the remaining die can take on, and there are $\binom{4}{3}=4$ ways to choose the die. There are $6$ ways that this can happen. Hence, $6\cdot 4\cdot5=120$ ways.

$\textbf{Case 2}$: Exactly four of the dices show the same value.

This can happen in $6$ ways.

Hence, the probability is $\frac{120+6}{6^{4}}=\frac{21}{216}\implies \frac{7}{72}\implies \boxed{\textbf{(B)}}$

Solution 3

We solve using PIE.

We first calculate the number of ways that we can have $3$ dice be the same and the other dice be anything. We therefore have $\binom{4}{3} \cdot 6 \cdot 6 = 144$ ways to have at least $3$ dice be the same.

But wait! We have overcounted the case where all $4$ dice are the same! Since the previous case occurs in each of these cases $4$ times, we must subtract the $4$-dice total three times in order to have them counted once. There are $6$ ways to have four dice be the same, so we our total count is $144 - 3(6) = 126$.

Therefore, our probability is $\frac{126}{6^4} = \boxed{\frac{7}{72}}$, which is answer choice $\boxed{\textbf{(B)}}$.

-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 $\frac{1}{6^4}$ chance that one number will be rolled four times in a row. Since there are six numbers on a die, we multiply by $6$ to see that the probability for the first case is $\frac{1}{6^3}.$


For the second case, consider the roll $AAAB$, where three of the dice are identical and the fourth differs. The probability of the first three rolling the same number is $1\cdot{\frac{1}{6}}\cdot{\frac{1}{6}},$ because the first number can be anything, and the second must be identical. The probability of the last roll being different is $\frac{5}{6}$, as it can be anything except for what has been previously rolled.


Multiplying these together, the probability for the second case is $\frac{5}{6^3}.$ However, there are $\frac{4!}{3!\cdot{1!}} = 4$ ways to arrange $AAAB$, so we must multiply by a factor of 4 to get the true probability for this case, which is $4(\frac{5}{6^3}) = \frac{20}{6^3}.$


Adding these two cases, we get the requested probability: $\frac{1+20}{6^3} = \frac{21}{216} = \frac{7}{72},$ or answer choice $\boxed{\textbf{(B)}}$


-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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_16&oldid=233112"