help with this nasty probability problem..?

Question

two blue marbles, two red marbles, and two green marbles are placed in a bucket. a player is asked to draw a marble.
> if the marble is blue, the player tosses 1 coin.
> if the marble is green, the player tosses 2 coins.
> if the marble is red, the player tosses 3 coins.
The player wins if the coin(s) all land tails.
what is the probablility of winning??

Answer 1

Equal probability for each marble, so...

1/3(1/2 + 1/4 + 1/8) = (1/3)*(7/8) = 7/24

Answer 2

P[blue] = P[red] = P[green] = 1/3.
P[win] = P[1 toss = tail | blue]*P[blue] + P[2 toss = tail | green]*P[green] + P[3 toss = tail | red]*P[red]
= 1/2 * 1/3 + 1/4 * 1/3 + 1/8 * 1/3 = 7/24.

Answer 3

p(win) =
p(all coins land tails) =
sum of [p(marble color)*p(all tails|marble color)] =
p(blue)*p(all tails|blue) + p(red)*p(all tails|red) + p(green)*p(all tails|green) =
(1/3)(1/2) + (1/3)(1/4) + (1/3)(1/8) =
(1/3)(7/8) =
7/24.

Answer 4

P = 1/3 * (0.5 + 0.5^2 + 0.5 ^3) = 29%