2006 AMC 10B Problems/Problem 21
Contents
[hide]Problem
For a particular peculiar pair of dice, the probabilities of rolling , , , , , and , on each die are in the ratio . What is the probability of rolling a total of on the two dice?
Solution
Let be the probability of rolling a . The probabilities of rolling a , , , , and are , , , , and , respectively.
The sum of the probabilities of rolling each number must equal 1, so
So the probabilities of rolling a , , , , , and are respectively , and .
The possible combinations of two rolls that total are:
The probability P of rolling a total of on the two dice is equal to the sum of the probabilities of rolling each combination.
Solution 2 (Not as bashy)
(Alcumus solution) On each die the probability of rolling , for , is There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs , , , , , and . Thus the probability of rolling a total of 7 is
Solution 3 (intuitive)
There are ways to get the sum of of the dice. Let's do case by case.
Case (Rolling a or a ): .
Case (Rolling a or a ): .
Case (Rolling a or a ): .
The rest of the cases are symmetric to these cases above. We have . We have . Therefore, our answer is
~Arcticturn
~Minor Edits by L314159265358979323846264
Solution 4 (cheap)
Notice that the ways to obtain a 7 are (6,1), (5,2), (4,3). Then, because the cases are symmetrical with (3,4), (2,5) and (6,1), look through the answer choices. You see a 4/63 and an 8/63, so obviously MAA wants you to not count the other symmetrical cases, thus giving the answer (C).
~mathboy282
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=3057
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=SBwVVADk1Nk ~David
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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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