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Probability Level 2

Suppose you have 10 balls in which 9 balls are of 1 kg and 1 of 0.5 kg. Find the minimum number of steps (number of times we have to compare the weights of the balls) using a beam balance in which we can guarantee finding out the ball weighing 0.5 kg

1 3 4 2

2 solutions

Calvin Lin Staff
Dec 3, 2014

[This is not a solution.]

The correct answer has been updated to 3.

Rayyan replied to the dispute in the following way:

We will make sets of 3 balls. We will place 2 sets on the beam balance.If they both are equal then we will take the 3rd set and please 2 balls from it on the beam balance. If they both are equal then then the third ball is the required ball otherwise if they are not equal then the one of lesser weight is the required ball. If in the the first step the 2 sets of ball are not equal in weight then we will repeat the same method with the set with lesser weight.

However, my concern is that this description only works for 9 total balls instead of 10. I believe that the answer is 3 weighings. Comments?

Agreed. The third set Rayyan refers to would in this case be composed of $4$ balls, and hence the suggested process would not guarantee identification of the lighter ball. One would need to divide this set of $4$ in half and either measure two versus two with a subsequent measurement of one versus one, or (potentially) make two measurements of one versus one. Thus in the worst case scenario we need to make $3$ measurements to guarantee identification of the lighter ball.

In general, if we have a two-pan balance and $n$ balls with one of the balls being either lighter or heavier than the others, and $3^{a} \lt n \le 3^{a+1}$ for some integer $a$ , then ( $a + 1)$ weighings will be required to identify the different ball. In this case $3^{2} \lt 10 \le 3^{3}$ , so the correct answer is $3$ .

Brian Charlesworth - 6 years, 6 months ago

Brian, sorry, I replied to Rayyan before seeing your comment, but no harm done. Ed Gray

Edwin Gray - 2 years, 8 months ago

This is a variant of a more difficult problem, which I have not submitted as a problem, because it asks to devise an algorithm, rather than supplying the number of times to weigh. The problem, which you may be familiar with, is known as the 12 ball problem. There are 12 balls, of which 11 are identical; however, 1 ball is slightly lighter or slightly heavier. The problem is to determine the odd ball, AND whether it is heavier or lighter using a balance scale three times; I think you will enjoy this if you are not familiar with it. Best Regards, Ed Gray

Edwin Gray - 2 years, 8 months ago

Ranik Chakraborty
Mar 29, 2015

put 5 balls each to both the pans...then take the balls of the pan weighing less and put them in pairs,if they weigh equal,then the left ball is the one with lesser weight...if they weigh unequal then place them in the pans and you can get the ball weighing less...hope i'm clear.

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