Determine the probability of getting 2 heads, 1 tail from flipping a coin three times.?

H = head and T = tail
P(H) = (1/2)
P(T) = (1/2)


Method 1 :
P(getting 2 H and 1 T from flipping a coin)
= C(3,2) × [P(H)]² × P(T)
= (3!/2!1!) × (1/2)² × (1/2)
= 3/8


Method 2 :
P(getting 2 H and 1 T from flipping a coin)
= P(HHT) + P(HTH) + P(THH)
= (1/2)×(1/2)×(1/2) + (1/2)×(1/2)×(1/2) + (1/2)×(1/2)×(1/2)
= 3/8

THH
HTH
HHT
P = 3/8

The are 4 possibilities, but the probability of each possibility is NOT equal to 1/4. P(2 heads 1 tail) = P(1 head 2 tails) = 3/8, while P(3 heads) = P(3 tails) = 1/8.

Does the order in which you get them matter?

If it does, there are 2*2*2=8 ways the flipping can result in. So 1/8 since 2 heads and 1 tail is one specific order.

If order doesn't matter, you can either get 3 heads (1), 2 heads 1 tail (2), 2 tails 1 head (3) or 3 tails (4) and since 2 heads 1 tail is one of those 4 possibilities, the probability is 1/4
Edit: Never mind the last one, there being three ways of 2 tails, 1 head slipped my head for some reason, my bad