For out come for the sum=2, only one possibility (1,1) and its probability is 1/36.
For out come for the sum=3, there are two possibilities (1,2) and (2,1)) and its probability is 2/36=1/18
For out come for the sum=4, there are 3 possibilities (1,3), (3,1) and (2,2) and its probability is 3/36= 1/12.
For out come for the sum=5, there are 4 possibilities
(1,4), (4,1), (2,3), (3,2) and its probability is 4/36=1/9.
For out come for the sum=6, there are 5 possibilities
(1,5), (5,1), (2,4), (4,2) and (3,3) and its probability is 5/36.
For out come for the sum=7, there are 6 possibilities
(1,6), (6,1), (2,5), (5,2) ,(3,4), (4,3) and its probability is 6/36.
For out come for the sum=8, there are 5possibilities
(2,6), (6,2) ,(3,5), (5,3),(4,4) and its probability is 5/36.
For out come for the sum=9, there are 4 possibilities
(3,6), (6,3), (4,5),(5,4)and its probability is 4/36.
For out come for the sum=10, there are 3 possibilities
(4,6),(6,4),(5,5)and its probability is 3/36.
For out come for the sum=11, there are 2 possibilities
(5,6), (6,5) and its probability is 2/36..
For out come for the sum=12, there is 1 possibilities
(6,6) and its probability is 1/36..
For out come for the sum=2,3,4,10,11,12, for each case the probability is less than 1/9. Answer.
2007-01-16 16:33:20
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answer #1
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answered by Anonymous
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We can do this the exhaustive way.
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
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
P(2) = 1/36
P(3) = 2/36 = 1/18
P(4) = 3/36 = 1/12
P(5) = 4/36 = 1/9
P(6) = 5/36
P(7) = 6/36 = 1/6
P(8) = 5/36
P(9) = 4/36 = 1/9
P(10) = 3/36 = 1/12
P(11) = 2/36 = 1/18
P(12) = 1/36
Just look for the values that are less than 1/9. The answer would be P(2), P(3), P(4), P(10), P(11), P(12)
2007-01-17 00:34:29
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answer #2
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answered by Puggy 7
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