A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume

Question

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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?

Mathematics

1 answer:

Leona [12.6K] · Answer 1 · 2026-04-07T20:08:09+03:00

Answer:

(a) 0.4096

(b) 0.64

(c) 0.7942

Step-by-step explanation:

The probability of the player winning is,

$P(W)=0.80$

Then, the probability of the player losing is,

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

The player engages in a video game with 4 different opponents.

It is stated that if the player is defeated by an opponent, the game ends.

The possible outcomes for the player to win are: {L, WL, WWL, WWWL and WWWW)

(a)

The outcomes from the four opponents are independent, meaning the result against one does not influence the results against the others.

The probability of the player defeating 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)

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

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

So, the probability of the player defeating at least two opponents in a game is 0.64.

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

The probability of the player defeating 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 of defeating all 4 opponents at least once in several games is,[TAG_89]]P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

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