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When we roll three 6-sided dice, what is the probability that the sum of the numbers rolled is at least 11?
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
E. \(1\)
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
E. \(1\)
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31 Oct 2022, 21:46
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av1901
The average value for each die is 3.5 (the median of the evenly spaced set: 1,2,3,4,5,6)
So, the average value for 3 dice is 3*3.5=10.5
Since the distribution of values is symmetrical above above and below the average, half of the permutations will be below 10.5, and half will be above 10.5.
Therefore, the probability is 50%, or 1/2
General Discussion
19 Jul 2025, 08:05
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What is the probability that the sum of the numbers rolled is at least 11?
The sum of the roll of three dice varies from (1,1,1) = 3 to (6,6,6) = 18
The number of cases for each sum will keep on increasing from 1 when the sum is 3 (1,1,1), till the time it becomes the maximum(27) and then will keep on decreasing till the time the number of cases for the sum is 1 again when sum is 18 (6,6,6)
Since, the sum goes from 3 to 18 and the pattern is symmetric so getting a sum from 3 to 10 (10-3+1 = 8 values for sum) has the same chance as getting the sum from 11 to 18 (18-11 + 1 = 8 values for sum)
=> Probability of Sum to be at least 11 (11 or more) = \(\frac{1}{2}\)
Sum of 3 dice rolls.jpg [ 63.62 KiB | Viewed 1023 times ]
Details of the cases below:
So, Answer will be D
Hope it helps!
Playlist on Solved Problems on Probability here
Watch the following video to MASTER Dice Rolling Probability Problems
The sum of the roll of three dice varies from (1,1,1) = 3 to (6,6,6) = 18
The number of cases for each sum will keep on increasing from 1 when the sum is 3 (1,1,1), till the time it becomes the maximum(27) and then will keep on decreasing till the time the number of cases for the sum is 1 again when sum is 18 (6,6,6)
Since, the sum goes from 3 to 18 and the pattern is symmetric so getting a sum from 3 to 10 (10-3+1 = 8 values for sum) has the same chance as getting the sum from 11 to 18 (18-11 + 1 = 8 values for sum)
=> Probability of Sum to be at least 11 (11 or more) = \(\frac{1}{2}\)
Attachment:
Sum of 3 dice rolls.jpg [ 63.62 KiB | Viewed 1023 times ]
Details of the cases below:
| Sum | Cases | Total Cases |
| 3 | (1,1,1) | 1 |
| 4 | (1,1,2) = 3 cases (2 can come in any of the three positions) | 3 |
| 5 | (1,1,3) = 3 cases (3 can come in any of the three positions), (1,2,2) = 3 cases (1 can come in any of the three positions) | 6 |
| 6 | (1,1,4) = 3 cases (4 can come in any of the three positions) , (1,2,3) - 3*2 = 6 cases, (2,2,3) = 1 case | 10 |
| 7 | (1,1,5) = 3 cases, (1,2,4) = 3*2 = 6 cases, (1,3,3) = 3 cases, (2,2,3) = 3 cases | 15 |
| 8 | (1,1,6) = 3 cases, (1,2,5) = 3*2 = 6 cases, (1,3,4) = 3*2 = 6 cases, (2,2,4) = 3 cases, (2,3,3) = 3 cases | 21 |
| 9 | (1,2,6) =3*2 = 6 cases, (1,3,5) = 2*3 = 6 cases, (1,4,4) = 3 cases, (2,2,5) = 3 cases, (2,3,4) = 3*2 = 6 cases, (3,3,3) = 1 case | 25 |
| 10 | (1,3,6) = 3*2 = 6 cases, (1,4,5) = 3*2 = 6 cases, (2,2,6) = 3 cases, (2,3,5) = 3*2 = 6 cases, (2,4,4) = 3 cases, (3,3,4) = 3 cases | 27 |
| 11 | (1,4,6) = 3*2 = 6 cases, (1,5,5) = 3 cases, (2,3,6) = 2*3 = 6 cases, (2,4,5) = 3*2 = 6 cases, (3,3,5) = 3 cases, (3,4,4) = 3 cases | 27 |
| 12 | (1,5,6) = 3*2 = 6 cases, (2,4,6) = 2*3 = 6 cases, (2,5,5) = 3 cases, (3,3,6) = 3 cases, (3,4,5) = 3*2 = 6 cases, (4,4,4) = 1 case | 25 |
| 13 | (1,6,6) = 3 cases, (2,5,6) = 3*2 = 6 cases, (3,4,6) = 2*3 = 6 cases, (3,5,5) = 3 cases, (4,4,5) = 3 cases | 21 |
| 14 | (2,6,6) = 3 cases, (3,5,6) = 3*2 = 6 cases, (4,4,6) = 3 cases, (4,5,5) = 3 cases | 15 |
| 15 | (3,6,6) = 3 cases, (4,5,6) = 3*2 = 6 cases, (5,5,5) = 1 case | 10 |
| 16 | (4,6,6) = 3 cases, (5,5,6) = 3 cases | 6 |
| 17 | (5,6,6) = 3 cases (5 can come in any of the three positions) | 3 |
| 18 | (6,6,6) | 1 |
So, Answer will be D
Hope it helps!
Playlist on Solved Problems on Probability here
Watch the following video to MASTER Dice Rolling Probability Problems
