In a bag there are 6 white, 4 black and 5 yellow balls. Three balls are taken at random from the bag. What is the probability that not all of them are of the same colour?

To solve this problem, we first need to calculate the total number of ways to draw 3 balls from the bag, and then calculate the number of ways to draw 3 balls of the same color. The difference between these two numbers, divided by the total number of ways to draw 3 balls, will give us the probability that not all of the balls are the same color.

Step 1: Calculate the total number of ways to draw 3 balls from the bag. This is a combination problem, because the order in which we draw the balls does not matter. The formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. In this case, n = 6 white + 4 black + 5 yellow = 15 balls, and k = 3 balls. So the total number of ways to draw 3 balls is C(15, 3) = 15! / [3!(15-3)!] = 455.

Step 2: Calculate the number of ways to draw 3 balls of the same color. This is the sum of the ways to draw 3 white, 3 black, or 3 yellow balls. For white, we have C(6, 3) = 6! / [3!(6-3)!] = 20. For black, we have C(4, 3) = 4! / [3!(4-3)!] = 4. For yellow, we have C(5, 3) = 5! / [3!(5-3)!] = 10. So the total number of ways to draw 3 balls of the same color is 20 + 4 + 10 = 34.

Step 3: Subtract the number of ways to draw 3 balls of the same color from the total number of ways to draw 3 balls, and divide by the total number of ways to draw 3 balls. This gives us the probability that not all of the balls are the same color: (455 - 34) / 455 = 421 / 455 = 0.9253, or 92.53%.