Table of contents |
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Problem Statement |
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Observations |
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Solution |
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Conclusion |
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In this puzzle, you have a bag with 20 red and 16 blue balls. You have to pull out two balls at a time, and if they are of the same color, you replace them with a blue ball. However, if they are of different colors, you replace them with a red ball. Once you take out the balls, you don't put them back in the bag, so the number of balls in the bag keeps decreasing. The question is, what color will be the last ball remaining in the bag?
To solve this puzzle, we need to make some observations. First, blue balls can only be reduced by two at a time, and only if you choose two blue balls. In no other way can you reduce the number of blue balls. On the other hand, red balls can be reduced by one in two situations: if you choose two different balls, or if you choose two red balls.
Now, since there are an even number of blue balls in the bag, they can only be reduced by two at a time. Therefore, at any point during the replacement process, you will end up with either 0 or 2 or 4... (even number) blue balls in the bag. There will never be an odd number of blue balls in the bag.
As a result, no matter what combination of replacements you choose, the last ball in the bag will always be a red ball.
In conclusion, the last ball remaining in the bag will be a red ball, no matter what combination of replacements you choose. This puzzle helps us understand the concept of probability and how certain conditions can lead to specific outcomes.