A bag contains 5 red, 4 blue and 3 green balls. Three

Solution

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.