We play a game with three bags of balls, where the game is best of five, and one wins when they select three winning balls. A bag is selected at the beginning of a game, and it does not change between selections within games. There are three bags described as follows, and selecting each is equally likely.
• The "lucky" bag, where selecting a winning ball has a 3/4 chance.
• The "impartial" bag, where selecting a winning ball has a 1/2 chance.
• The "unlucky" bag, where selecting a winning ball has a 1/4 chance.
Answer the following questions:
- What is the probability that the game ends in the first three consecutive draws?
This seemed to mean you select three winning balls in a row or three losing balls. By the law of total probability:$$ P(\text{Game ends in 3 draws}) = \frac{1}{3}(\frac{3^3}{4^3} + \frac{1^3}{2^3} + \frac{1^3}{4^3}) + \frac{1}{3}(\frac{1^3}{4^3} + \frac{1^3}{2^3} + \frac{3^3}{4^3}) = \frac{3}{8}$$
- What is the probability that the you lose the game?
This seems to be the probability that you draw exactly three losing balls (regardless of the game ending before five balls are drawn, but this is where I'm confused). Again by the law of total probability:
$$ P(\text{Lose the game}) = \frac{1}{3}(\frac{1^3}{4^3} + \frac{1^3}{2^3} + \frac{3^3}{4^3}) = \frac{3}{16}$$
- What is the probability that you selected one winning ball and three losing balls?
By the law of total probability:
$$ P(\text{1 winning ball, 3 losing balls}) = \frac{1}{3}(\frac{1^3}{4^3}\cdot\frac{3}{4} + \frac{1^4}{2^4} + \frac{3^3}{4^3}\cdot\frac{1}{4})$$
- Given that you selected one winning ball and three losing balls, what is the probability that you were drawing from the "lucky" bag?
By the definition/formula for conditional probability, seems like this could be $$\frac{1}{4^3}\frac{3}{4}$$ divided by the answer to the part right above (i.e. the probability that we draw one winning ball and three losing ones).