Probability of non-broken lamps We have two boxes of lamps (A, B), each including 50 lamps such that box A has 45 non-broken lamps and box B has 43 non-broken lamps. In each step, we choose one of the boxes with equal probability and take one lamp out of it. We do this three times. What is the probability that all these three picked lamps are non-broken?
I personally think that as there are 8 different possibilities for the sequence of lamps to be chosen, therefore, we can calculate the probability of each of these conditions and sum them up, but I think I may be wrong.

There is indeed 8 possibilities of box combination, and you can weight im by there probability to be chosen and then add up. That's what the law of total probability tells you. 
You have an eight change to end in each of the cases. So you get:
$$P(E)= 1/8 \cdot P(non-broken | \text{box} A,A,A\text{ was chosen}) +1/8 \cdot P(non-broken | \text{box} B,A,A\text{ was chosen}) +..$$. I would use some symmetry, to avoid calculating all 8. Is this enough for you to proceed?
