statistics help!!?

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

P(1tail) = 1/2
P(2tails) = 1/4
P(3tails) = 1/8

P(winning) = [1/2 + 1/4 + 1/8] /3 = 7/8 * 1/3 = 7/24 = 29.16666%

Because each color marble is equally likely (2/6), we can just take the average of the three probabilities listed above. If each color marble had a different probability, then you would have to use a weighted average.

*Sorry about my previous error...I must have hit a wrong button on my calculator.

Answer 2

The previous poster's idea is correct but his numbers are wrong. (1/2) + (1/4) + (1/8) is 7/8, not 23/24.

The correct answer is (2/6)*(1/2) + (2/6)*(1/4) + (2/6)*(1/8)
That's (1/3) * (7/8), or 7/24. In decimal form, it's about 0.2917.

Answer 3

There is 1/3 chance to pick a blue marble, 1/3 chance to pick a green marble, and 1/3 chance to pick a red marble.

Break down probably to win by color of marble

Blue: 1/2 chance to win (only two outcomes, 1 winning)

Green: TT (win), TH, HT, HH = 1/4 chance to win (four outcomes, 1 winning)

Red: TTT (win), TTH, THH, THT, HTT, HTH, HHT, HHH
(eight outcomes, 1 winning) = 1/8 chance to win

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Note: 1 coin toss, chance of winning was 2^-1 (1/2)
2 coins tossed, chance of winning was 2^-2 (1/4)
3 coins tossed, chance of winning was 2^-3 (1/8)
this is the quick way to determine the chance of winning without writing down every possible outcome, for a binary coin toss.
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Chance to pick blue (1/3) and win (1/2) = 1/3 * 1/2 = 0.167
Chance to pick green (1/3) and win (1/4) = 1/3 * 1/4 = 0.083
Chance to pick red (1/3) and win (1/8) = 1/3 * 1/8 = 0.042

Total chance to win = 0.167 + 0.083 + 0.042 = 0.292
Total chance to win = 29.2 % <-- ANSWER