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

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We split this problem into 2 cases.
 
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 <math>1 x \dfrac{1}{6} x \dfrac{1}{6} x \dfrac{1}{6} = \dfrac{1}{216}</math>
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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>
 
Second, we calculate the probability that three are the same and one is different. Adding these up, we get <math>\dfrac{7}{72}</math>, or <math>\boxed{\textbf{(B)}}</math>.
 
Second, we calculate the probability that three are the same and one is different. Adding these up, we get <math>\dfrac{7}{72}</math>, or <math>\boxed{\textbf{(B)}}</math>.
  

Revision as of 16:25, 20 February 2014

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

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

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