Let N be the total number of possible outcomes when three dice are thrown.
Each die has 6 faces, so the total number of outcomes is
N=6×6×6=216.
Let
S be the sum of the numbers on the three dice.
The minimum possible sum is
1+1+1=3.
The maximum possible sum is
6+6+6=18.
We need to find the probability that the sum S is a prime number. The prime numbers between 3 and 18 (inclusive) are:
3, 5, 7, 11, 13, 17.
Now we need to count the number of ways to obtain each of these prime sums. Let
(d1,d2,d3) be the numbers on the three dice. The order matters (e.g.,
(1,1,2) is different from
(1,2,1)).
Sum = 3:
The only combination is
(1,1,1).
Number of ways: 1 .
Sum = 5:
Possible combinations (listing partitions in non-decreasing order to avoid duplicates, then counting permutations):
(1,1,3) : Permutations:
‌=3 ways
((1,1,3),(1,3,1),(3,1,1))(1,2,2) : Permutations:
‌=3 ways
((1,2,2),(2,1,2),(2,2,1))Number of ways:
3+3=6.
Sum = 7:
Possible combinations
(d1≤d2≤d3) :
(1,1,5):3 ways
(1,2,4):3!=6 ways
(1,3,3):3 ways
(2,2,3):3 ways
Number of ways:
3+6+3+3=15.
Sum = 11:
Possible combinations
(d1≤d2≤d3) :
(1,4,6):3!=6 ways
(1,5,5):3 ways
(2,3,6):3!=6 ways
(2,4,5):3!=6 ways
(3,3,5):3 ways
(
3,4,4 ): 3 ways
Number of ways:
6+3+6+6+3+3=27.
Sum = 13:
Possible combinations
(d1≤d2≤d3) :
(1,6,6):3 ways
(2,5,6):3!=6 ways
(3,4,6):3!=6 ways
(
3,5,5 ): 3 ways
(
4,4,5 ): 3 ways
Number of ways:
3+6+6+3+3=21.
Sum = 17:
Possible combinations
(d1≤d2≤d3) :
(5,6,6):3 ways
(Note: sums like
(4,x,y) would give max
4+6+6=16, so 5 must be present)
Number of ways: 3 .
Now, we sum the number of ways for each prime sum to get the total number of favorable outcomes:
Total favorable outcomes
=1+6+15+27+21+3=73.
The probability of getting a sum that is a prime number is the ratio of favorable outcomes to the total possible outcomes:
‌ Probability ‌=‌| ‌ Number of favorable outcomes ‌ |
| ‌ Total number of outcomes ‌ |
=‌.
