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P(6) = 5/36 = 0.138888... ≈ 13.9%

When two fair dice are rolled, the sample space (all possible outcomes) has 36

equiprobable events:(1,1), (1,2), (1,3), (1,4), (1,5),(1,6), (2,1), (2,2), (2,3), (2,4),

(2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5),

(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

There are 5 events where the sum of the dots of each die is equal to 6:

(1,5), (2,4), (3,3), (4,2), (5,1).

So in two fair dice are rolled the probability that the total number of spots shown is equal to 6 is, P(6) = 5/36 = 0.138888... ≈ 13.9%

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Q: If two fair dice are rolled what is the probability that the total number of spots shown is equal to 6?
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If two twelve side fair dice are rolled what it the probability that the total number of spots shown is equal to 5?

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Related questions

If two fair dice are rolled what is the probability that the total number of spots shown is equal to 15?

If two fair dice were rolled, there would be 36 outcomes. (1,1),(1,2),....,(6,6) The maximum sum would be 12. Therefore, the probability that the total number of spots shown is equal to 15 is zero.


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