A bag contains 12 marbles, of which 4 are red, 4 are blue, and 4 are green. Marbles are drawn randomly in triplets (three at a time) without replacement until the bag is empty. The probability that every drawn triplet contains exactly one marble of each colour is :

Step-by-step solution

The total number of ways to divide 12 marbles into 4 unordered triplets is: Total ways = 12! (3!)^4 4! = 15400 For each triplet to contain exactly one marble of each colour, we can fix the 4 red marbles in 4 distinct groups. Then, we distribute the 4 blue marbles among these groups in 4! ways, and the 4 green marbles in 4! ways. Favorable ways = 4! 4! = 24 24 = 576 Thus, the required probability is: P = 576 15400 = 72 1925 Alternatively, using conditional probability for sequential drawing of triplets: P₁ = ⁴C₁ ⁴C₁