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 1

Answer:

(a) The PMF of X is: $P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let X = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable X follows a Geometric distribution.

The probability mass function of X is:

$P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....$

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of X is:

$P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

              = 1 - P (X = 1) - P (X = 2)

              $=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64$

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

$E(X)=\frac{1}{p}$

Compute the expected number of opponents contested in a game as follows:

$E(X)=\frac{1}{p}=\frac{1}{0.20}=5$

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

              = 1 - P (X = 1) - P (X = 2) - P (X = 3)

              $=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512$

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

$E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2$

Thus, the expected number of game plays until a player contests four or more opponents is 2.