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

m
(Solution 3)
(4 intermediate revisions by the same user not shown)
Line 27: Line 27:
  
 
Hence, the probability is <math>\frac{120+6}{6^{4}}=\frac{21}{216}\implies \frac{7}{72}\implies \boxed{\textbf{(B)}}</math>
 
Hence, the probability is <math>\frac{120+6}{6^{4}}=\frac{21}{216}\implies \frac{7}{72}\implies \boxed{\textbf{(B)}}</math>
 +
 +
==Solution 3==
 +
 +
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.
 +
 +
But wait! We have overcounted the case where all <math>4</math> dice are the same! Since the previous case occurs in each of these cases <math>4</math> times, we must subtract the <math>4</math>-dice total three times in order to have them counted once. There are <math>6</math> ways to have four dice be the same, so we our total count is <math>144 - 3(6) = 126</math>.
 +
 +
Therefore, our probability is <math>\frac{126}{6^4} = \boxed{\frac{7}{72}}</math>, which is answer choice <math>\boxed{\textbf{(B)}}</math>.
 +
 +
-FIREDRAGONMATH16
  
 
==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}}

Revision as of 05:23, 18 May 2021

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

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

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

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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