This problem asks for the minimum number of balls we need to take out from a bag to be absolutely sure we have at least one ball of each colour (red, blue, and green). This type of problem involves thinking about the worst-case scenario.

Balls in the Bag

First, let's list the number of balls of each colour in the bag:

• Red balls: 8
• Blue balls: 10
• Green balls: 17

The total number of balls in the bag is \(8 + 10 + 17 = 35\).

Worst-Case Scenario for Minimum Balls

To guarantee getting at least one ball of each colour, we must consider the scenario where we try to avoid getting one of each colour for as long as possible. The worst possible outcome is that we draw all the balls of the colours that have the largest numbers first.

The colours with the largest quantities are blue (10 balls) and green (17 balls).

In the worst case, we would first draw all the blue balls and all the green balls before drawing a single red ball.

Number of blue balls = 10

Number of green balls = 17

Total balls of the two most numerous colours = \(10 + 17 = 27\)

After drawing these 27 balls, the bag would contain only red balls. We would have drawn all 10 blue balls and all 17 green balls.

Guaranteeing One of Each Colour

Once we have drawn all 27 balls from the two most numerous colours, the very next ball we draw must be a ball of the third colour, which is red in this case.

So, after drawing 27 balls (all blue and all green), the next ball we draw guarantees that we have at least one red ball, in addition to the blue and green balls we already drew.

Therefore, the minimum number of balls required to ensure getting at least one ball of each colour is the sum of the two largest quantities plus one.

Minimum number of balls = (Number of blue balls + Number of green balls) + 1

Minimum number of balls = \(10 + 17 + 1\)

Minimum number of balls = \(27 + 1\)

Minimum number of balls = \(28\)

Drawing 28 balls guarantees that we will have at least one red ball, one blue ball, and one green ball.

Let's verify this with the options:

• Option 1: 19 (Not enough in the worst case)
• Option 2: 18 (Not enough in the worst case)
• Option 3: 28 (Matches our calculation)
• Option 4: 27 (Exactly the number of balls of the two most numerous colours; drawing 27 could mean we got all blue and all green, but no red)

The minimum number of balls needed is 28.