- #1
anyalong18
- 4
- 0
Two fair dice are rolled. What is the probability of rolling a sum that exceeds 3?
Probability of Rolling Sum > 3 with Two Dice
- Context: MHB
- Thread starter anyalong18
- Start date
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- Tags
- Dice Probability Rolling Sum
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Discussion Overview
The discussion revolves around calculating the probability of rolling a sum greater than 3 when two fair dice are rolled. It includes considerations of possible outcomes and methods for determining the probability.
Discussion Character
- Mathematical reasoning
- Technical explanation
Main Points Raised
- One participant asks for the probability of rolling a sum that exceeds 3.
- Another participant suggests that the probability might be 1/36.
- A clarification is made that a sum exceeding 3 is equivalent to a sum greater than or equal to 4.
- A detailed enumeration of all possible sums from rolling two dice is provided, noting that there are 36 total outcomes.
- A question is raised about how many outcomes yield a sum of 4 or higher, indicating that this number will be used to calculate the probability.
- One participant proposes an alternative method of counting the outcomes that result in sums of 2 or 3 and subtracting that from the total outcomes.
Areas of Agreement / Disagreement
Participants have not reached a consensus on the probability value, and multiple approaches to the problem are presented, indicating ongoing discussion and exploration of the topic.
Contextual Notes
The discussion does not resolve the exact number of outcomes that exceed a sum of 3, nor does it clarify the implications of the proposed methods for calculating the probability.
- #2
skeeter
- 1,103
- 1
- #3
anyalong18
- 4
- 0
is it 1/36skeeter said:
- #4
skeeter
- 1,103
- 1
a sum that exceeds 3 is a sum $\ge$ 4
- #5
HOI
- 921
- 2
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.
- #6
HOI
- 921
- 2
You might find it simpler to count the number of rolls that give "2" or "3" and subtract that from 36.
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