Of the 6 drawings, the red marbles can be drawn any of the 6 
times, but the probability of drawing the red ones in any of 
the 6 times is the same as the probability of drawing all the 
red ones in the beginning, so we calculate that probability 
and multiply by the number of choices of times to draw the red 
ones, which is the combinations of 6 drawings, choose the number 
of reds to draws.  

P(drawing 1 red, 5 greens) = 
6C1*P(drawing the red one 1st) =
6C1(5/10)(5/9)(4/8)(3/7)(2/6)(1/5) = 1/42

P(exactly 2 reds, 4 greens) = 
6C2*P(drawing the red ones 1st & 2nd) =
6C2(5/10)(4/9)(5/8)(4/7)(3/6)(2/5) = 5/21

P(exactly 3 reds, 3 greens) = 
6C3*P(drawing the red ones 1st, 2nd & 3rd) =
6C3(5/10)(4/9)(3/8)(5/7)(4/6)(3/5) = 10/21

P(exactly 4 reds, 2 greens) = 
6C4*P(drawing the red ones 1st, 2nd, 3rd & 4th) =
6C4(5/10)(4/9)(3/8)(2/7)(5/6)(4/5) = 5/21

P(exactly 5 reds, 1 green) = 
6C3*P(drawing the red ones 1st, 2nd, 3rd 4th) =
6C5(5/10)(4/9)(3/8)(2/7)(1/6)(5/5) = 1/42

Event:        
number          Ex-
 of     Prob    pect-
 red     of      a-
ones   event   tions
------------------------
  n      p      n*p
  1     1/42    1/42
  2     5/21   10/21
  3    10/21   30/21
  4     5/21   20/21
  5     1/42    5/42
------------------------
       sum=1   sum=3=expected number of red marbles

Edwin