Aroom has 3 lamps. from a collection of 12 light bulbs of which 5 are not working, richard selects 3 at random and puts them in the sockets. what is the probability that he will have light?

I calculated
(7∗6∗5)÷(12∗11∗10)=.159
or 15.9%
But that's for selecting 3 good bulbs. Do the same with two bulbs
(7∗6)÷(12∗11)=
or 31.8%
And with one bulb
7÷12=.583
or 58.3%So the answer seems to depend on the criteria for "having light", which is what makes it so confusing. Plus, we'd like to consider the probability of any of these conditions obtaining in any combination. The process yields errors if we keep trying to figure out how to choose a good bulb.
So let's turn the question on its head and ask what's the probability of no light?
( 12
  3  )=12!÷3!9!=20
then how many ways to choose 3 bad bulbs?
(12
   5 )=12!÷5!7!=792

Divide and subtract from one to get the probability of there being light. 1 - 20/792 = 97.5%