A bowl contains 10 red balls, 5 blue balls, and 8 green balls. A woman selects balls at random without looking in the bowl.
How many balls must she select to be sure that she has at least one ball of each color?
To be sure that she has at least one ball of each color, the woman needs to keep selecting balls until she has picked all the different colors in the bowl.
Let's calculate the worst-case scenario, assuming she selects the maximum number of balls of the same color each time.
Since there are three different colors (red, blue, and green), the woman needs to pick at least one ball of each color.
In the worst-case scenario, she starts by picking all the red balls (10), then all the blue balls (5). Now, there are still two colors remaining (red and blue), so she needs to keep picking balls until she selects a green ball to have one ball of each color.
Considering that she has already selected 15 balls (10 red + 5 blue) and there are 8 green balls remaining in the bowl, she would need to pick at least 8 more balls to ensure she gets a green ball.
Therefore, she must select a total of 15 + 8 = 23 balls to be sure that she has at least one ball of each color.