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)
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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.