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

(→‎Solution: Added second half of solution, finished)
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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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 +
==Solution 2==
 +
 +
Note that there are two cases for this problem
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 +
Case 1: Exactly three of the dices show the same value.
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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.
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Case 2: Exactly four of the dices show the same value.
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This can happen in <math>6</math> ways.
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Hence, the probability is <math>\frac{120+6}{6^{4}}=\frac{21}{216}\implies \frac{7}{72}\implies \boxed{\textbf{(B)}}</math>
  
 
==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 14:27, 12 June 2017

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 $1 * \dfrac{1}{6} * \dfrac{1}{6} * \dfrac{1}{6} = \dfrac{1}{216}$.

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

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

Solution 2

Note that there are two cases for this problem

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.

Case 2: Exactly four of the dices show the same value.

This can happen in $6$ ways.

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

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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