Using C(n,r) = n!/((n-r)!r!) below...
There areoutcomes possible. So 8 will be the denominator.
How many outcomes have 0 heads? 1 (TTT)
How many outcomes have 1 head? C(3,1) = 3 (HTT, THT, TTH)
How many outcomes have 2 heads? C(3,2) = 3 (HHT, HTH, THH)
How many outcomes have 3 heads? C(3,3) = 1 (HHH)
Thus the distribution is:
0 heads: 1/8
1 head: 3/8
2 heads: 3/8
3 heads: 1/8
(Note they sum to 8/8 = 1, which is expected since one of these outcomes must occur)
The function notation is P(r heads) =,