Question

. A player of a video game is confronted with a series of opponents and has an 75% 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 that a player defeats at least two opponents in a game? (b) What is the expected number of opponents contested in a game? (c) What is the probability that a player contests four or more opponents in a game? (d) What is the expected number of game plays until a player contests four or more opponents?

Answer 1

Solution:

A player of a video game is confronted with a series of opponents and has an 75% probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents.

This follows a bernoulli distribution with P =0.75 , Q = 1 - 0.75 = 0.25

a)
The player defeats at least 2 opponents in a game is
P(3<= x <= Infinity ) = 1 - P(X<=2) = SUM ( 0.75)^ (x-1) (0.25) = 0.5625

b)
Expected mean = 1 / p = 1/0.75 = 4

c)
P(4 < = X < = Infinity) = 1 - P(1<=X<=3) = SUm ( 0.75)^(x-1) ( 0.25) = 0.4218