Difference between revisions of "2014 AMC 10B Problems/Problem 16"
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==Solution== | ==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 <math>1 x \dfrac{1}{6} x \dfrac{1}{6} x \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>. | ||
==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 15:24, 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?
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 
Second, we calculate the probability that three are the same and one is different. Adding these up, we get 
, or 
.
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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