Light bulb probability

Question

We have two lamps that only function if both have a functioning bulb in them. We also have 5 functioning and 5 non-functioning light bulbs in a drawer. How should we proceed in trying out the light bulbs, if we want to use the lowest amount possible to light the lamps up?

Answer 1

The best I've been able to do is $12$ changes, but I don't know if that's the minimum. Here's my algorithm. In all cases, I assume the lamp doesn't light.

1. AB (2 changes)
2. AC
3. BC
4. DE (2 changes)
5. DF
6. EF At this point, we know there is at most one good bulb among A,B,C and at most one good bulb among DEF, so there are at least three good bulbs among G,H,I,J. So we test GH and if that fails, IJ is bound to succeed, giving four additional changes at most.

Answer 2

Another way to get minimum $12$ changes. Try $3$ pairs (6 changes): $$AB, CD, EF$$ They can be double defective or mixed.

Then, the rest $2$ pairs ($GH,IJ$) must have at least $2$ normal light bulbs. Try: $$GH \ (2 \text{ changes})\\ GI \ (1 \text{ change})\\ GJ \ (1 \text{ change})$$ If they don't light, then: $$HJ \ (2 \text{ changes})$$ will definitely light.