Question

PRINTER VERSION NEXT I BACK ES Problem 2,152 A player of a video game is confronted with a series of 4 opponents and a(n) 82 % 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). Round your answers to 4 decimal places. (a) What is the probability that a player defeats all 4 opponents in a game? (b) What is the probability that a player defeats at least 2 opponents in a game? (c) If the game is played 3 times, what is the probability that the player defeats all 4 opponents at least once? Click if you would like to Show Work for this question: pen Show Work

Answer 1

Solution: Probability of defeating an opponent = p = 0.82 Number of opponents in a game = n = 4 Let variable x denotes the number of opponents defeated in a game. Part-A: Probability that a player defeats all 4 opponents = P(x = 4) Using Binomial distribution: P(x = 4) = f(4) =4C4(0.82)4 x (1 – 0.82) 4-4 = 0.4521 Hence, P(x = 4) = 0.4521 Therefore, probability that a player defeats all 4 opponents = 0.4521 Part-B: Probability that a player defeats at least 2 opponents = P(x > 2) Using Binomial distribution: P(x = 2) = f(2)=4C2 x (0.82)2 x (1 -0.82)4-2 = 0.1307 P(x = 3) = f(3) = 4C3 x (0.82)3x (1 – 0.82)4–3 = 0.3970 P(x = 4) = f(4) = 4C4 x (0.82)4 x (1 – 0.82)4-4 = 0.4521 Hence, P(x > 2) = 0.1307 + 0.3970+ 0.4521 = 0.9798 Therefore, probability that a player defeats at least 2 opponents = 0.9798 Part-C: As calculated in Part-A above, probability of defeating all 4 opponents in a game = 0.4521 Number of times the game is played = n = 3 Let variable x denotes the number of games in which a player defeats all 4 opponents. Probability that all 4 opponents are defeated at least once = P(x > 1) Using Binomial distribution: P(x = 1) = f(1) = 3C1 x (0.4521)1x (1 – 0.4521)3-1 = 0.4071 P(x = 2) = f(2) = 3C2 x (0.4521)2 x (1 – 0.4521)3-2 = 0.3360 P(x = 3) = f(3) = 3C3 x (0.4521)3 x (1 – 0.4521)3-3 = 0.0924 Hence, P(x > 1) = 0.4071 + 0.3360+ 0.0924 = 0.8355 Therefore, probability that all 4 opponents are defeated at least once = 0.8355