anyalong18
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Two fair dice are rolled. What is the probability of rolling a sum that exceeds 3?
Probability of Rolling Sum > 3 with Two Dice
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skeeter
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anyalong18
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is it 1/36skeeter said:
skeeter
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a sum that exceeds 3 is a sum $\ge$ 4
HOI
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When rolling two dice the possible sums are 1+ 1= 2 to 6+ 6= 12. The specific possible outcomes are
1+ 1= 2
1+ 2= 3
1+ 3= 4
1+ 4= 5
1+ 5= 6
1+ 6= 7
2+ 1= 3
2+ 2= 4
2+ 3= 5
2+ 4= 6
2+ 5= 7
2+ 6= 8
3+ 1= 4
3+ 2= 5
3+ 3= 6
3+ 4= 7
3+ 5= 8
3+ 6= 9
4+ 1= 5
4+ 2= 6
4+ 3= 7
4+ 4= 8
4+ 5= 9
4+ 6= 10
5+ 1= 6
5+ 2= 7
5+ 3= 8
5+ 4= 9
5+ 5= 10
5+ 6= 11
6+ 1= 7
6+ 2= 8
6+ 3= 9
6+ 4= 10
6+ 5= 11
6+ 6= 12
A total of 6x6= 36 outcomes, not all different.
Now, how many of those "exceed 3" (i.e. are 4 or higher)? The probability of exceeding 3 is that number divided by 36.
1+ 1= 2
1+ 2= 3
1+ 3= 4
1+ 4= 5
1+ 5= 6
1+ 6= 7
2+ 1= 3
2+ 2= 4
2+ 3= 5
2+ 4= 6
2+ 5= 7
2+ 6= 8
3+ 1= 4
3+ 2= 5
3+ 3= 6
3+ 4= 7
3+ 5= 8
3+ 6= 9
4+ 1= 5
4+ 2= 6
4+ 3= 7
4+ 4= 8
4+ 5= 9
4+ 6= 10
5+ 1= 6
5+ 2= 7
5+ 3= 8
5+ 4= 9
5+ 5= 10
5+ 6= 11
6+ 1= 7
6+ 2= 8
6+ 3= 9
6+ 4= 10
6+ 5= 11
6+ 6= 12
A total of 6x6= 36 outcomes, not all different.
Now, how many of those "exceed 3" (i.e. are 4 or higher)? The probability of exceeding 3 is that number divided by 36.
HOI
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You might find it simpler to count the number of rolls that give "2" or "3" and subtract that from 36.