You have 20 blue balls and 14 red balls in a bag. You put your hand in and remove 2 at a time. If they’re of the same color, you add a blue ball to the bag. If they’re of different colors, you add a red ball to the bag. (assume you have a big supply of blue & red balls for this purpose. note: when you take the two balls out, you don’t put them back in, so the number of balls in the bag keeps decreasing). What will be the color of the last ball left in the bag?
Once you tackle that, what if there are 20 blue balls and 13 red balls to start with?
See Solution : Balls in a bagThere are 3 possible cases of removing the two balls…
a) If we take off 1 RED and 1 BLUE, in fact we will take off 1 BLUE
b)If we take off 2 RED, in fact we will take off 2 RED (and add 1 BLUE)
c) If we take off 2 BLUE, in fact we will take off 1 BLUE
So In case of (a) or (c), we are only removing one blue ball, but we always take off red balls two by two.
1) 20 Blue, 14 Red balls
If there are 14 (even) number of red balls, we can not have one single red ball left in the bag, so the last ball will be blue.
2) 20 Blue, 13 Red balls
Now as the no. of red balls is odd, there will be one single red ball in the bag with other blue balls, and whenever we remove 1 red and 1 blue ball, we end up taking off only the blue ball. So the red ball will be the last ball in the bag.
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Got it
Can anyone explain me how to remain red for 20blue and 14 red
i think the solution is the other way round i.e for 20,14 its red and for 20,13 its blue.
Please verify and let me know if i am wrong
sorry this is correct ,my bad