There are 3 boxes, one of them contains red balls, another contains green balls and the last contains red balls and green balls.

Question

There are 3 boxes, one of them contains red balls, another contains green balls and the last contains red balls and green balls. These boxes are labelled with the following labels: "Red balls" "Green balls" "Red balls and green balls". Knowing that all the labels are placed incorrectly, what's the minimum amount of extractions ( taking a ball from any box without looking at what is inside ) required to determine the content of each box?

Answer 1

We know that the labels MUST BE WRONG. Thus we can establish relevant cases to solve this problem.

For notation, I will use r() for the label "RED", g() for the label "GREEN" and rg() for the label "REd&GREEN." Inside the parentheses will be the actual contents of the box. Thus r(g) would be a box labelled red, but the box only contains green balls.

There are 6 ways the boxes can be labelled, we know this from a basic understanding of combinatorics that with three unique items, they can be placed 3! ways, or 3*2*1=6

We can eliminate any combinations where any of the labels are correct (ie r(r) or rg(rg))

Case 1: r(r) g(g) rg(rg)

Case 2: r(r) g(rg) rg(g)

Case 3: r(g) g(r) rg(rg)

Case 4: r(rg) g(r) rg(g)

Case 5: r(g) g(rg) rg(r)

Case 6: r(rg) g(g) rg(r)

Eliminate cases 1,2,3, and 6 because each of these cases have at least one correct label (bolded).

Thus the incorrect labelling can only be either case 4 or case 5.

We know the boxes have a wrong label, thus we can be strategic about our pulls, as we know there is only one of two ways these can be labelled. Go to the rg() bin, as it is guaranteed to be only whatever you pull on your first try (the problem with the r() and g() bins is that there is risk of having two colors and incorrectly identifying the label on your first pull. Remember, we want to minimize our pulls to get the correct answer.

If, from the box labelled rg() you pull a green ball, you know that it must be case 4, and you know the true identity of the boxes.

If, from the box labelled rg() you pull a red ball, you know that it must be case 5, and you know the true identity of the boxes.

Thus, you only need to pull ONE ball to determine the content of each box.

This was a particularly fun question to answer! If you need any help or clarification please reach out and I will explain further.

Answer 2

Let the boxes be A, B and C labelled as follows:

[R] [G] [R&G]

A B C

Extraction #1: We take 1 ball from box C

If this ball is red:

box C has all red balls

box B has the mix of red and green balls

box A has all green balls

If this ball is green:

box C has all green balls

box A has the mix of red and green balls

box B has all red balls

Therefore, only 1 extraction is required.

Answer 3

You only need to pick one ball

Each box is labeled incorrectly, so we know the box labeled "Red / Green" either contains only Red balls or only Green balls. Pick one ball from it

1) If the ball you picked is Red, then this is the box with only red balls. Then the box labeled Red must contain only Green balls (because otherwise there would be a correct label) and the box labeled Green must contain both Red and Green balls

2) If the ball you picked is Green, then this is the box with only Green balls. Then the box labeled Green must contain only Red balls and the box labeled Red must contain both Red and Green balls

Answer 4

It depends on how many balls are in the mixed box...

FOr example, if there are 3 red and 2 green, you would have to

sample 4.

If there are 6 green and 2 red, then you have to sample 7.

It is POSSIBLE, you can get away with only having to sample 2 of them,

if you are lucky... one red and one green