Imagine that you have three boxes, one containing two black balls, one containing two white balls, and the third containing one black ball and one white ball.
The boxes were originally labelled for their contents (BB - WW - BW) but someone has inadvertently switched the labels so that now every box is incorrectly labelled.
Without looking inside, you are allowed to take one ball at a time out of any box that you wish, and by this process of sampling, you are to determine the contents of all three boxes.
What is the smallest number of drawings needed to do this?
Since every box is mislabeled we have two possibilities.
#1:(label-contents):(BW-BB),(BB-WW),(WW-BW)
#2:(label-contents):(BW-WW),(BB-BW),(WW-BB)
Either way by choosing one ball from the box labeled BW, we determine whether that first box has both white balls or both black balls. If the BW labeled box has WW, then the BB labeled box must have BW contained for it cannot have BB in it.
A similar argument can be made for the other possibility. Thus, the lowest number of drawings is one.
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