(7 votes, average: 4.29 out of 5)You have three bags and three labels. One bag has only red balls, one has only blue balls and one has both red and blue balls. Three labels are R, B and RB. R label was meant for the bag with only red balls, B label was meant for the bag with only blue balls and RB for the bag with both red and blue balls. Ram by mistake labelled the bags wrongly such that all the labels are wrong, how many minimum number of balls he should pick and from what bags to correct the labels? It is given that each bag has unlimited number of balls to be picked.
See Solution : Red and blue ballsHow ?
As the label of all the bags are wrong, bag with label RB can either have all red balls or all blue balls.
Now ram will either get red ball or the blue ball.
Case 1: Ram gets the red ball
Bag with label RB is the bag with all the red balls(R).
Bag with label B can either be RB or R(as all the labels are wrong) but as we already know R bag so it can only be RB.
We know 2 bags correctly so third bag with label R should be B.
Case 2: Ram gets the blue ball
Bag with label RB is the bag with all the blue balls(B).
Bag with label R can either be B or RB(as all the labels are wrong) but as we already know B bag so it can only be RB.
we know 2 bags correctly so third bag with label B should be R.
I can think of the solution as: “3, where 1 is from each bag”.
Explanation:
Draw 1 ball from each bag. You will get 2 of one colour and 1 of the other colour. Whichever bag you drew only 1 colour of, contains only that colour. Swap that label with the appropriate bag. So one bag is now correctly labelled. Since we are given that all bags are wrongly labelled at the start, it means the remaining 2 bags are still wrongly labelled (since 1 label was not touched in the swap, and we can’t have only 1 wrong label – it has to be 2 or 0 wrong). Swap their labels, and all bags are now correctly labelled.