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?

Question

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?

anonymous · Answer

key parts of the question:
- it is a series of 4 opponents (not "all 4 at once")
- results are independent (player's character doesn't get tired)

anonymous · Answer

first one is straight forward. to beat them all you multiply and get $.8^4$

anonymous · Answer

at least 2 means 2 or 3 or 4, so you can calculate those three or calculate 
beats 0 or 1 and subtract that result from 1
either way

anonymous · Answer

Thank you so much for ur help

anonymous · Answer

do you know how to compute the probability you lose all game or the probability you win exactly one?

anonymous · Answer

losing all is easy, it is $(.2)^4$ 
losing 1 is $4\times (.2)^3\times (.8)$
compute those two, add them and then subtract the total from 1