So this is a homework question and I'm looking for guidance if I have done the question correctly. Box 1 contains 1000 bulbs of which 10% are defective. Box 2 contains 2000 bulbs of which 5% are defective. Two bulbs are picked from a randomly selected box. What is the probability that both bulbs are defective.
So the probability that you can pull a defective bulb from box 1 is 1/10. The probability of pulling a defective bulb from box 2 is 1/5. Since we are pulling two bulbs this experiment is completed twice. So for 1 experiment (1st bulb pulled) I did the cartesian product for that probability giving 1/50. And for the second experiment I just took that same probability and calculated the cartesian product again giving 1/2500.
Suppose the box picked is the first then the probability of two defective is $p_{1,2}=(1/10)\times (1/10)$. If you had picked the second box the probability would have been $p_{2,2}=(1/20)\times (1/20)$.
As the box is picked at random each has a 50% chance of being picked so the final probability of getting two defectives is: $$ p_2=0.5 \times p_{1,2}+0.5 \times p_{2,2} $$
Note here I am assuming we can ignore the change in probability of choosing a defective after the first pick. This seems justified given the vague wording rather than the more precise 1000 (or 2000 for the second box) bulbs of which 100 (or 100 ...) are defective.
If we change the wording so that box 1 contains 1000 of which 100 are defective and box 2 contains 2000 of which 100 are defective we get $p_{1,2}=(100/1000)\times (99/999)$ and $p_{2,2}=(100/2000)\times (99/1999)$ ...
The first method gives $p_2\approx 0.00625$ and the second $p_2\approx 0.00619$