Question 488990
Good that you tried.
For something this small, it's fine just to list all possibilities, but this should be done systematically to avoid missing or double counting.
I.e., HHH HHT HTH HTT THH THT TTH TTT (count H=0 and T=1 and this is counting in binary from 0 to 7).
Possibly because your list was random, you "didn't count perfectly".
1. Has 3 hits, P = 3/8
2. Has 4 hits, P = 1/2
3. Is exactly like 2 because H and T are symetric. P = 1/2
4. It is impossible to get any probability out of this example = 1/6; P = 7/8.

There is indeed a better way to do this.
Note that none of the 4 questions care about the order, which is reasonable since the coins were dropped all at once.
Thus you can expand (H+T)^3 [works for any "3", e.g., 5 coins --> (H+T)^5]
HHH + 3HHT + 3HTT + TTT  (not in any coin order, just H written before T).
Meaning: There is only one way to get all H; 3 ways to get HHT = 3 ways to pick which coin is the T.
From this you just look [and add]:
1. 2nd term: 3 --> 3/8
2. 3rd and 4th terms: 3+1 = 4 --> 1/2
3. 1st and 2nd terms: 4
4. 1st 3 terms: 1+3+3 = 7
Yes, this is "classic probability" although I haven't heard that term.