How about we consider our specimen space. We have 3 coins, each with 2 conceivable results. This gives us 2^3=8 particular results.
Ω={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}Ω={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
We realize that the coins are reasonable, so it can be seen that each of these results is similarly likely. That is, every individual grouping has a likelihood equivalent to one over the span of the specimen space, or 1/8.
Presently we can just tally. There are 4 successions that incorporate no less than 2 heads:
{HHH,HHT,HTH,THH}{HHH,HHT,HTH,THH}
The likelihood of any of these occasions happening is equivalent to the entirety of the individual probabilities of these occasions happening, so the likelihood of no less than two heads is 4/8 or 1/2.