SOLUTION: 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 (an
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Question 418107: 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?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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)
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Binomial Problem with n = 4 and p = 0.8
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(a) What is the probability that a player defeats all four opponents?
Ans: 0.8^4 = 0.4096
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(b) What is the probability that a player defeats at least two of the opponents in a game?
Ans: 1 - P(x = 0 or x=1)
= 1 - [0.2^4 + 4C1(0.8)*0.2^3]
= 1 - [0.0016 + 4*0.0064]
= 1 - [0.0080]
= 0.992
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(c) If the game is played three times, what is the probability that the player defeats all four opponents at least once?
P(defeat all at least once out of 3) = 1 - P(losses all three)
= 1 = 0.02^3
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I'll leave that to you.
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Cheers,
Stan H.
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