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Suppose X = sum

Pr(X = 3n where n is an integer or X>4)

= 1 - Pr(X ≠ 3n and X ≤ 4)

= 1 - pr(X = 2 or X = 4) since these are the only two outcomes that meet the requirements of the event.

= 1 - [Pr(X=2) + Pr(X=4)]

= 1 - [1/36 + 3/36]

= 1 - 4/36 = 1 - 1/9

= 8/9

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Q: When two fair dice is rolled there are 36 possible outcomes what is the probability that the sum is a multiple of 3 or greater than 4?
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