A bag contains 5 red, 4 blue and 3 green balls. Three balls are drawn at random from the bag one after another without replacement. What is the probability that among the three balls drawn, exactly two are of the same colour and the third is of a different colour?

Total balls = 5 + 4 + 3 = 12 Total number of ways to choose 3 balls out of 12 = C(12, 3) = 220 We want: exactly two of one colour and the third of a different colour. Count favourable cases by splitting into cases based on which colour forms the pair. Pair of Red balls (5 red), and one of another colour 2 Red and 1 Blue: C(5, 2) × C(4, 1) = 10 × 4 = 40 2 Red and 1 Green: C(5, 2) × C(3, 1) = 10 × 3 = 30 Total for Red pair = 40 + 30 = 70 Pair of Blue balls (4 blue), and one of another colour 2 Blue and 1 Red: C(4, 2) × C(5, 1) = 6 × 5 = 30 2 Blue and 1 Green: C(4, 2) × C(3, 1) = 6 × 3 = 18 Total for Blue pair = 30 + 18 = 48 Pair of Green balls (3 green), and one of another colour 2 Green and 1 Red: C(3, 2) × C(5, 1) = 3 × 5 = 15 2 Green and 1 Blue: C(3, 2) × C(4, 1) = 3 × 4 = 12 Total for Green pair = 15 + 12 = 27 Total favourable outcomes = 70 + 48 + 27 = 145 Required probability = Favourable / Total = 145 / 220 = (divide by 5) → 29 / 44 So, the probability is 29/44.