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Solved: A player of a video game is confronted with a
Chapter 3, Problem 126E(choose chapter or problem)
A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents.
(a) What is the probability mass function of the number of opponents contested in a game?
(b) What is the probability that a player defeats at least two opponents in a game?
(c) What is the expected number of opponents contested in a game?
(d) What is the probability that a player contests four or more opponents in a game?
(e) What is the expected number of game plays until a player contests four or more opponents?
Questions & Answers
QUESTION:
A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents.
(a) What is the probability mass function of the number of opponents contested in a game?
(b) What is the probability that a player defeats at least two opponents in a game?
(c) What is the expected number of opponents contested in a game?
(d) What is the probability that a player contests four or more opponents in a game?
(e) What is the expected number of game plays until a player contests four or more opponents?
ANSWER:Step 1 of 5
Given that the player has 80% probability of defecting each player
Then \(p=1-0.8=0.2\)
Let \(X\) be the no.of opponents in the game
a) We have to write the probability mass function of the no.of opponents in the game
Here \(X\) is following the geometric distribution
The pmf of geometric distribution is \(P(x)=p(1-p)^{x-1} ; x=1,2,3, \ldots\)
Hence the probability mass function of the no.of opponents in the game is
\(P(x)=(0.2)(0.8)^{x-1}\)