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In a bag of 6 white marble 4 black marbles and 1 red marble what is the least number of marbles that you need to take out of the box to ensure that you get 3 marbles of the same color?
Wiki User
∙ 11y ago
8 i think 8 i think I guess it's 6.. and my point is that you have to try it through the worst sequence ever. starting with the red one (the one that is impossible for you to get another one, and will never reach the three same colored wanted) then start getting one of each of the two other colors each time, alternating, so it will take longer for you to have several of the same color: example: red > black > white > black > white > black (3 black ones)
or red > white > black > white > black > white (3 white ones) or (just to prove that alternate sequence isn't needed) white > black > black > red > white > (...)
(impossible to get any marble without getting the third of it's color, and by not getting the red one, this would have been impossible to reach, so to be SURE, you have to count this one in the first five, or it would come or not in the 5th position)
in this case, a longer sequence would have got the three equal-colored marbles some time ago. remembering, that this sequence is just to explain that 6 would do the job, and these balls would have been caught in a handful of marbles =) ps.:sorry for my poor English.
Wiki User
∙ 11y ago
You would need to take out at least 5 marbles to ensure you have 3 of the same color. This is because in the worst-case scenario, you could end up with 2 marbles of each color after drawing 4 marbles (2 white, 2 black) and adding 1 more marble guarantees that you will have at least 3 of one color.
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== ==
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