Answer:
(a) 0.4096
(b) 0.64
(c) 0.7942
Step-by-step explanation:
The probability that the player wins is,

Then the probability that the player losses is,

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](https://tex.z-dn.net/?f=P%28W%29%5Ctimes%20P%28W%29%5Ctimes%20P%28W%29%5Ctimes%20P%28W%29%3D%5BP%28W%29%5D%5E%7B4%7D%20%3D%280.80%29%5E%7B4%7D%3D0.4096)
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) = 
Thus, the probability that the player defeats at least two opponents in a game is 0.64.
(c)
Let <em>X</em> = 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 
The probability distribution of binomial is:

The probability that the player defeats all the 4 opponents at least once is,
P (<em>X</em> ≥ 1) = 1 - P (<em>X</em> < 1)
= 1 - P (<em>X</em> = 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](https://tex.z-dn.net/?f=%3D1-%5B%7B3%5Cchoose%200%7D%280.4096%29%5E%7B0%7D%20%281-0.4096%29%5E%7B3-0%7D%5D%5C%5C%3D1-%5B1%5Ctimes1%5Ctimes%20%280.5904%29%5E%7B3%7D%5C%5C%3D1-0.2058%5C%5C%3D0.7942)
Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.