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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) .

AWhat is the probability that a player defeats all four opponents?.
BWhat is the probability that a player defeats at least two of the opponents in a game?

Verified Answer :

toporc
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
meerkat18

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 0.8^20.2^2 + 4C3 0.8^3*0.2^1 + 0.4096 = 0.9728

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