Three fair dice are thrown. What is the probability of getting a total of 15 given that they exhibit three different numbers that are in arithmetic progression?

Step-by-step solution

Let the three numbers on the dice be x, y, z . The condition is that x, y, z are in arithmetic progression and are distinct. Let the numbers be a-d, a, a+d where d > 0 . Since the numbers are between 1 and 6 , we have 1 a-d and a+d 6 . For d=1 : (a-1, a, a+1) can be (1,2,3), (2,3,4), (3,4,5), (4,5,6) . There are 4 sets. Each set has 3! = 6 permutations. Total outcomes = 4 6 = 24 . For d=2 : (a-2, a, a+2) can be (1,3,5), (2,4,6) . There are 2 sets. Each set has 3! = 6 permutations. Total outcomes = 2 6 = 12 . Total