A player of a video game is confronted with a series of opponents and has an 80%

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. What is the probability mass function of the number of opponents contested in a game? What is the probability that a player defeats at least two opponents in a game? What is the expected number of opponents contested in a game? What is the probability that a player contests four or more opponents in a game? What is the expected number of game plays until a player contests four or more opponents?

Explanation / Answer

this is a geomterric distribution

a)pmf of number of opponents contested=P(X=x) =P(untill X-1 contest win and xth contest loses)=(0.8)x-1*(0.2)

b)probabilty that at least 2 opponents defeated =(0.8)2=0.64

c)expected number =1/0.2=0.5

d)probability that a player contests four or more opponents in a game =P(at least he defeats 3 opponents)

=(0.8)3=0.512

expected number of games =1*0.2+2*0.8*0.2+3*0.8*0.8*0.2=0.904

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