A box contains 21 ribbons out of which [A] ribbons are white, and rest is green in colour. The probability of drawing a white ribbon and a green ribbon together is 18/35. Now, [B] green ribbons are added in the box such that the probability of drawing a green ribbon from the box is 1/2. Now, [C] yellow ribbons are added to the box such that the probability of drawing a yellow ribbon from the box is 1/5. Find the value of C-B?

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