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Difference between revisions of "2014 AMC 10B Problems/Problem 16"
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==Solution==
 
==Solution==
  
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We split this problem into 2 cases.
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We split this problem into <math>2</math> cases.
  
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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 * \dfrac{1}{6} * \dfrac{1}{6} * \dfrac{1}{6} = \dfrac{1}{216}</math>.
+
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 <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 * 1 * \dfrac{1}{6} * \dfrac{1}{6} * \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.
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<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.
  
 
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
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 +
==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 18:28, 15 September 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

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

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