Please help with this probability

Night reader wrote:The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least two of the tosses the coin will turn up tails?

the probability that the coin is toosed 3 times =(1/2)^3=1/8
there are 4 combination of tossing time that at least of the tosses the coin will turn up tails.
TTH
THT
HTT
TTT ( becase its "at least" probability, we will have to consider that all time its tossed is tail)
so the final probability is 4*1/8=1/2
is 1/2 the right answer?

Rahul, thanks for the solution. Below I have my questions related to different interpretation of the probabilities listed, and I am including the official answer and solution which I don't understand at all, as it operates with the method I am unaware

Possible exactly two tails: (TTH), (THT) and (HTT)
Each have a probability = (1/2)*(1/2)*(1/2) = 1/8

Can I say, P(not Tail-One Head)=1/2=P(two heads) above => 1/2 means P(One Head) which is enough for us => then 1/8 is equivalent to P(TTH) & P(THT) & P(HTT)

we cancel this

Probability of getting exactly two tails= 3*(1/8) = 3/8

Also, below P(Head)=0 and we must be consistent with the probability tree here, if we start with P(Head) we finish with P(Head)

Possible exactly three tails: (TTT)
Probability of getting exactly three tails = (1/2)*(1/2)*(1/2) = 1/8

Therefore, Probability of getting at least two tails = (3/8) + (1/8) = 4/8 = 1/2

summing up two probabilities P(TTH) & P(THT) & P(HTT) + P(TTT) we get 1/2 + 0 = 1/2


And below is the official explanation of which I don't understand the red color selected part, please shed some light on this

Explanation: The words "at least" signal that you'll be working with more than one probability, then adding them together later. In this case, those multiple probabilities are the probability that two of the three tosses turn up tails, and the other is the probability is that all three of the tosses turn up tails. The only ways the coin tosses can be arranged to make one of those things possible:
H T T
T H T
T T H
T T T
There are four possible outcomes, in the …rest three of which two tosses turn up tails, and in the last of which all three turn up tails. The total number of possibilities is the product of the number of possibilities for each toss: (2)(2)(2) = 8
Probability is, as always, the number of desired outcomes divided by possible outcomes: 4/8 = 1/2

Thank you Rahul. I find your method as more expletive and concept backed. I will keep it for notes to tackle other probs.