Question

A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume that the results from opponents are independent (and that when the player is defeated by an opponent the game ends).
A. What is the probability that a player defeats all four opponents in a game?
B. What is the probability that a player defeats at least two opponents in a game?
C. If the game is played three times, what is the probability that the player defeats all four opponents at least once?

Answer 1

Answer:

(a) 0.4096

(b) 0.64

(c) 0.7942

Step-by-step explanation:

The probability that the player wins is,

$P(W)=0.80$

Then the probability that the player losses is,

$P(L)=1-P(W)=1-0.80=0.20$

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = $P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096$

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = $1-P(L)-P(WL)$

                                    $=1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64$

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

(c)

Let X = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable $X\sim Bin(n=3, p=0.4096)$

The probability distribution of binomial is:

$P(X=x)={n\choose x}p^{x} (1-p)^{n-x}$

The probability that the player defeats all the 4 opponents at least once is,

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             $=1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942$

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.