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
