- #1

swmmr1928

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## Homework Statement

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. The player continues to contest opponents until defeated.

What is the probability that a player defeats at least two opponents 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?

## Homework Equations

f(x)=(1-p)^(x-1)*p

E=1/p

## The Attempt at a Solution

I know that these are Bernoulli trials.

I chose Geometric distribution because the number of 'trials' is not fixed.

Defeat at least two opponents:

pmf(1)+pmf(2)=cmf(2)=0.2+0.13=0.36

Contest four or more:

1-cmf(3)=1-[pmf(3)+pmf(2)+pmf(1)]=1-0.488=0.512

Expected games to contest four or more:

1/0.512=1.9?This is an illogical answer.