# Statistics-Probability Distribution of Discrete Random Variable

1. Sep 15, 2012

### swmmr1928

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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:

2. Sep 15, 2012

### LCKurtz

You might start by stating what your random variable $X$ represents. Although I can guess, you should tell us because there are a couple of ways the geometric distribution is set up and it needs to be clear to both of us.

3. Sep 15, 2012

### swmmr1928

X represents the number of opponents faced per set of trials.

4. Sep 16, 2012

### LCKurtz

You haven't shown enough calculation to follow what you did. I suspect you may be using the wrong formula for $f(x)$. Are you using $f(x) = .8^{x-1}\cdot 2$? So to defeat at least 2 opponents you want $P(X \ge 3) = 1 - P(X=1)-P(x=2)$. I get $.64$ for that one.