A box contains 21 ribbons out of which [A] ribbons are

Solution

Let number of white ribbons in the box is ‘x’. Number of green ribbons in the box = (21 – x) According to question, [x × (21 – x)]/ ²¹C2 = 18/35 x (21 – x)/210 = 18/35 x (21 – x) = 18 × 6 x² – 21x + 108 = 0 x = 12 or x = 9 So, the box contains either 12 white ribbons and 9 green ribbons or 9 white ribbons and 12 green ribbons. Now, if ‘B’ green ribbons are added and probability of drawing a green ribbon from the box becomes 1/2, it means number of green ribbons was less than number of white ribbons in the box initially. So, the numbers of white and green ribbons in the box initially were 12 and 9 respectively. Therefore, [A] = 12 and [B] = 3. Let number of yellow ribbons added in the box be ‘z’. So, [z/ (21 + 3 + z)] = 1/5 5z = z + 24 4z = 24 z = 6 So, [C] = 6 Required difference = 6-3 = 3