A bag contains 6 red balls, 5 green balls and 4 yellow balls. How many balls must you take out to be sure that you have at least 3 balls of each color?
I literally have no idea for this question. Should I just put 15(take all of them out to be sure?) or is there some kind on formula that I'm missing out on. THANKS!! You don't have to do it for me, but if there's a formula or something maybe tell me the name of it? Thanks everyone, Samantha.
You might take all 6 red balls, all 5 green balls first. So, you might have to take out 6+5+3=14 balls to be sure you also have 3 yellows.
Extra credit: how many do you need to be sure you have 3 balls of some color? Hint: There are 3 colors.
To determine the minimum number of balls you need to take out in order to be sure that you have at least 3 balls of each color, you can use the Pigeonhole Principle. The Pigeonhole Principle states that if you have more objects (pigeons) than you have categories (pigeonholes) to put them in, then at least one category must contain more than one object.
In this scenario, the different colors of balls represent the categories, and you want to find the minimum number of balls that guarantees you have at least 3 balls of each color.
To solve the problem, consider the worst-case scenario, where you initially draw the maximum number of balls of two colors (let's say red and green). To ensure you have at least 3 balls of each color, you need to draw balls of the third color (yellow).
In the worst-case scenario, you can draw all 6 red balls and all 5 green balls, leaving only the yellow balls in the bag. This means you have taken out a total of 11 balls.
To guarantee you have at least 3 balls of each color, you need to draw 2 more yellow balls. Thus, taking into account the worst-case scenario, you need to take out a minimum of 11 + 2 = 13 balls.
So, the answer is you must take out at least 13 balls to be sure that you have at least 3 balls of each color.