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?

Answer 1

(a) $P(4\ out\ of\ 4)=0.8^{4}=0.4096$

(b)
$P(2\ out\ of\ 4)=4C2\times 0.8^{2}\times0.2^{2}=0.1536$
$P(3\ out\ of\ 4)=4C3\times0.8^{3}\times0.2=0.4096$
P(4 out of 4)=0.4096
P(at least 2) = 0.1536 + 0.4096 + 0.4096 = 0.9728

Answer 2

It is given here that there is an 80% probability of defeating each opponent. Hence if four opponents are present, the probability that a player defeats all four opponents is equal to nCm p^n q^m or 4C4 0.8^4* 0.2^0 equal to 0.4096. b. the probability that a player defeats at least two of the opponents in a game is4C2 0.8^2*0.2^2 + 4C3 0.8^3*0.2^1 + 0.4096 = 0.9728