Respuesta :
Answer:
(a) 0.4096
(b) 0.64
(c) 0.7942
Step-by-step explanation:
The probability that the player wins is,
[tex]P(W)=0.80[/tex]
Then the probability that the player losses is,
[tex]P(L)=1-P(W)=1-0.80=0.20[/tex]
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) = [tex]P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096[/tex]
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) = [tex]1-P(L)-P(WL)[/tex]
[tex]=1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64[/tex]
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 [tex]X\sim Bin(n=3, p=0.4096)[/tex]
The probability distribution of binomial is:
[tex]P(X=x)={n\choose x}p^{x} (1-p)^{n-x}[/tex]
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)
[tex]=1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942[/tex]
Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.
