By the way, it is not clear to me from the wording of the question if h, t, t is the same as t, h , t or not
the first way there are 6 possible outcomes. The second way there are 8
If order does not matter
1 h h h
2 h h t
3 h t t
4 t t t
probability of each event here is 1/4
if order does matter
1 h h h
2 h h t
3 h t h
4 h t t
5 t t t
6 t t h
7 t h t
8 t h h
probability of each event here is 1/8
In the second list, listing all possible outcomes there is one way to get three heads p(3h) = 1/8
three ways of getting two heads
p(2h) = 3/8
three ways of getting 1 head
p(1h) = 3/8
one way of getting 0 heads
p (0h) = 1/8
let's look at a binomial distribution where p(h) = 1/2
n = number of trials = 3
k = number of successes (number of heads)
p(h = k) = C(n,k) p^k (1-p)^k
p(h=3) = c(3,3)(1/2)^3(1/2)^0
= 3!/[3!*0!](1/8)(1) = 1/8
p(h=1) = C(3,1)(1/2)(1/2)^2
c(3,1) = 3!/[1!(3-1)!] = 3*2/2 = 3
p(h=1) = 3(1/2)(1/4) = 3/8
you will find that p(h=2) = C(3,2)((1/2)^2(1/2)^1 = 3/8
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