Question

(ii) the sum of the numbers on their upper faces
is at the least 6.​

Answer 1

If it’s a dice so the problem of sum of numbers on upper faces is atleast 6 means sum >= 6 but <=12
So favourable chances = (3,3),(3,4),(3,5),(3,6),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(4,2),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
So total favourable events = 21
Probability = 21/36