Question 1203531
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Using  C(n,r) = n!/((n-r)!r!)   below...


There are {{{ 2^3 = 8 }}} outcomes 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) =  {{{ C(3,r)/8  }}},  {{{0<=r<=3}}}