Question 873974: Two fair dice are thrown, find the probability of getting:
1= A double six
2= A doublet
3= Sum of two dice will be eleven
4= The sum is at least 8
5= Sum of number is less than 6
6= Sum of number is 7.
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! There are  or  possible outcomes
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
1) 
2) 
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Here are all the sums using the same order as above
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
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3)
4)
5)
6)
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website! 36 possible outcomes: P = Particular/ALL possible
P(2 6s) = 1/36
P(double) = 6/36 = 1/6
P(sum 11) = 2/36 = 1/18
etc: Count them...
___1_2_3__4__5__6
1|_2_3_4__5__6__7
2|_3_4_5__6__7__8
3|_4_5_6__7__8__9
4|_5_6_7__8__9_10
5|_6_7_8__9_10_11
6|_7_8_9_10_11_12
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