"A bowl has 100 pieces of colored candy. 48 green, 30 red, 12 yellow and 10 blue. They are all wrapped in foil, so you do not know the color of any piece of candy. Write the least number of pieces you must take to be certain that you have at least 15 pieces of the same color?" -- MOEMS (Math Olympiad) problem

So again we have a worst case scenario problem (or Pigeonhole Principle problem) here. For my previous worst case scenario problem video explanations click here

How to solve the problem:

So we need 15 pieces of same colored candy. If we are lucky we can pick 15 candies and they would all turn up to be the same color. But alas that's not always the case. So what's the worst that can happen here?

We can remove all the blue and yellow candies and we would still not have 15 candies of the same color. Next, we can remove 14 green candies and 14 red candies from the bowl and still we would not have what we want!

But the next candy we remove will be a green or red candy (those are the only ones left in the bowl in case you're wondering) and that will make it 15 candies of the same color (either green or red). So how many candies did we have to remove?

12 yellow

10 blue

14 green

14 red

plus the last one. So we had to remove 51 candies to get 15 of the same color!!

Try solving this problem on your own now -

"Tucker has a jar containing 4 green and 5 blue marbles. What is the minimum number of marbles Tucker needs to remove from the jar, without replacement, to guarantee that he has at least four marbles of the same color among those removed?" -- Math Counts 2016 School Round.

Post your answer in the comments below.

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