SOLUTION: Suppose that three fair dice are thrown. Compute the probability of the following events: (a) the event is �big�, that is, the sum of the three dice is at least 10 but the three d

Question 945151: Suppose that three fair dice are thrown. Compute the probability of the following events:
(a) the event is �big�, that is, the sum of the three dice is at least 10 but the three dice
do not show the same value;
(b) the event that the sum is 13;
(c) the event that a specific combination of two different numbers appears, e.g. {3, 4}
appears.

Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!
Suppose that three fair dice are thrown. Compute the probability of the following events:
(a) the event is �big�, that is, the sum of the three dice is at least 10 but
the three dice do not show the same value;

Hmm! Does this mean that none of the three are the same? Or can 2 be the same
and the other different?  I will assume none of the three are the same.

The largest roll must be 5 or 6, because 4+3+2 is only 9.

If the largest roll is a 6, the other two must be at least 4.
The other two can be any of these 9 sets:

{1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5} 

If the largest roll is a 5, the other two must be {1,4}, {2,3}, {2,4}, {3,4}

That's 13 sets of 3 rolls with sum at least 10 where all are different.
There are 3! = 6 permutations of each.
So there are 13*6 or 78 "big" rolls.

So the probability of "big" is 76/216 = 19/54

(b) the event that the sum is 13;

There are only 2 sets of rolls that sum to 13. {6,5,2} and {6,4,3}
Each can be permuted 3!=6 ways, so there are 2*6 = 12 rolls of 13. 

So the probability is 12/216 = 1/18

They are {3,4,1},{3,4,2},{3,4,3},{3,4,4},{3,4,5},{3,4,6}

4 of them are all different and can be permted 3! or 6 ways.
That's 4*6 = 24 ways.
The other 2, {3,4,3} and {3,4,4} can be permuted 3 ways each
That's 2*3 = 6 ways.
Total = 24+6 = 30

Probability = 30/216 = 5/36

Edwin

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