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Black and White (Posted on 2003-02-25) Difficulty: 2 of 5
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?

See The Solution Submitted by Ravi Raja    
Rating: 3.1250 (8 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution Comment 25 of 25 |
Smallest number of drawings  required to do this is ONE.

We draw precisely  1 ball from the incorrectly labelled  BW jar.
Since it is incorrectly  labelled, the BW jar either contains  two black balls or two white balls.   ............(i)
CASE-1: The ball drawn from BW jar is white.
Then, by (i), the other ball must be white. So, this jar is relabelled as WW.
Therefore, BB jar cannot  contain two white balls. Also, being incorrectly  labelled it cannot contain 2 black balls. Threfore, it must contain precisely  1  black and 1 white ball. So, this jar is relabelled  as BW. 
The remaining  jar, that is WW must then be relabelled  as BB.
CASE-2: THE ball drawn from BW jar is black.
Then, following arguments  similar to Case-1, the BW jar should be relabelled as BB.
The WW jar should be relabelled as BW, and:
BW jar should be relabelled as WW.

  Posted by K Sengupta on 2021-12-31 23:28:26
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