2014 AMC 10B Problems/Problem 16

Revision as of 16:37, 20 February 2014 by DivideBy0 (talk | contribs) (Solution: Added second half of solution, finished)

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 $\boxed{\textbf{(B)}}$.

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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