In one urn lie two white balls, in another two black ones, in the third - one white and one black. Each urn had a sign indicating its contents: WW, BB, WB. Someone switched the signs so that now each sign incorrectly indicates the contents of the urn. It is allowed to draw a ball from any urn without looking inside. What is the minimum number of draws needed to determine the composition of all three urns?

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Substantially use the condition that NONE of the labels correctly indicate the contents.

## Solution

Take a ball from the urn labeled БЧ. If we draw a white ball, then this urn contains two white balls (otherwise, the label would correctly indicate the contents). Next, we determine that the urn labeled ЧЧ can only contain one white and one black ball (otherwise, the label would correctly indicate the contents). The only remaining possibility is that the urn labeled ББ contains two black balls. If, however, we draw a black ball from the urn labeled БЧ, then similarly, we determine that it contains two black balls, the urn labeled ББ contains one white and one black ball, and the urn labeled ЧЧ contains two white balls.

## Answer

only one draw is sufficient!