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?

Answer 1

Answer:

a) $P(X=x)=f(x)=p^{x-1}q=p^{x-1}(1-p)$

b) 0.64

d) 5

c) 0.4096

Step-by-step explanation:

Hi!

Lets say for example that the player defeats 7 oponents and the 8th defeats him. The probability of this, as the battles are independent, is:

$P(X=8)=0.8^7*0.2$

We are calling X to the number of opponents contested in the game. We can generalize to:

$P(X=x)=f(x)=p^{x-1}q=p^{x-1}(1-p)$

p = probability of success

f(x) = probabiliy mass function

To answer b) and d) we need to know P(X ≥ 2) and  P(X ≥ 4). So lets calculate the general case:

$P(X\geq i)=1- (1-p))\sum_{j=0}^{i-1}p^j =p^i$

We are using the well know geometric series (for r<1):

$\sum_{i=0}^{n-1}r^i=\frac{1-r^n}{1-r}$

Then:

b) $P(X\geq 2)=0.8^2=0.64$

d)$P(X\geq 4)=0.8^4=0.4096$

c) The expected value of X is:

$E(X) =\sum_{x=1}^{\infty} xf(x) = (1-p) \sum_{x=1}^{\infty}xp^{x-1}=(1-p)\frac{1}{(1-p)^2}=\frac{1}{1-p}$

$E(X)=\frac{1}{0.2}=5$

We are using another well know result:

$\sum_{x=1}^{\infty}xp^{x-1}= \frac{1}{(1-p)^2}$