Question 325275
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Successful outcomes are 3, 6, 9, 10, 11, 12.  2, 4, 5, 7, and 8 don't count because none of them are either multiples of 3 or greater than 8. (8 is just 8, NOT greater than 8).


To make a 3 you need 1 & 2 or 2 & 1, so 2 ways.


To make a 6 you need 1 & 5, 2 & 4, 3 & 3, 4 & 2, or 5 & 1, so 5 ways.


You get to figure out the number of ways to make each of 9, 10, 11, and 12.


Once you have determined the number of ways to make each of the allowable combinations, add them all up to make the numerator of your probability fraction. You already have the denominator, i.e. the number of possible outcomes.


John
*[tex \Large e^{i\pi}\ +\ 1\ =\ 0]
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