Statistics/ probability... I dont understand this 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

This is a question where you get to use
-conditional probability (the probability that event B occurs, given that event A already has occured),
-sequences of independant events:
P(A AND B) = P(A) * P(B)
and
-complements:
P(not A) = 1 - P(A)

P(neither A nor B) = (1-P(A))*(1-P(B))
P(A or B) = 1 - P(neither A nor B)
or, if you look for an exclusive or (meaning only one of A or B:
P(only one of A and B) = 1 - P(A and B) - (P(neither A nor B))

and so on.

Do a "decision tree" of all possible outcomes: begin with
Blue marble (P = 2/6) and 1 coin lands tail (P= 1/2)
Overall probabilities: Win= 1/6, lose = 1/6

Green marble (P=2/6), and 2 coins both on tail (P = 1/4)
Overall probabilities: Win = 1/12 Lose = 1/4

Red marble (P = 2/6), and 3 coins all on tail (P=1/8)
Win = 1/24 Lose = 7/24

Put everything on a common denominator (24)

Condition Win Lose
blue marble 4/24 4/24
green marb 2/24 6/24
red marble 1/24 7/24
Total = 24/24

Win = 7/24
Lose = 17/24

Answer 2

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.

Answer 3

Seems like a straight-forward problem to me. What part don't you understand?

Marbles are small round balls about 1 cm to 3 cm in diameter. They come in various colors.

A bucket is a thing that you can put marbles and other things into.

Are there other parts of the problem you don't understand?

If you mean that you don't understand how to SOLVE the problem, then please ask that question.