Independent Condition Probability Problem

1. Aug 30, 2006

Giuseppe

Hello, I had this problem to do for statistics. I am pretty sure I did the first two parts correct. The third, im not so sure about. I'd appreciate and guidance.

A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume that the results from opponents are independent (and that when the player is defeated by an opponent the game ends
(a) What is the probability that a player defeats all four opponents?
(b) What is the probability that a player defeats at least two of the opponents in a game?
(c) If the game is played three times, what is the probability that the player defeats all four opponents at least once?

My work:
(a) Since the probability of beating a player is 80 % and is also independent, then the probability of winning a game is .08 x .08 x .08 x .08
which comes out to .4096

(b) Since here the player only needs to beat two opponents so .08 x .08
which comes out to .64

(c) I am not too sure here, any help?

2. Aug 30, 2006

e(ho0n3

(a) looks correct. (b) unfortunately is not that simple. If I remember correctly how to do this correctly, the answer is the sum of the probability of defeating two, three and four opponents.

(c) uses a similar strategy to (b).