A player of a video game is confronted with a series of 3 opponents and a(n) 78%

Question

A player of a video game is confronted with a series of 3 opponents and a(n) 78% 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 3 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 2 times, what is the probability that the player defeats all 3 opponents at least once?

Explanation / Answer

Solution:

(a)

A player of game is confronted with a series of 3 opponents.

Probability of defeating one opponent is 0.78.
Compute the probabilityof defeating all three opponents.

Required Probability P(X = 3) = (0.78)3 = (0.78)(0.78)(0.78) = 0.4745

(b)

If the player loses once, he stops playing. Therefore, to win 2 games, he must win the first 2 games and lose on the third. The probability of that occuringis (0.78)(0.78)(0.22) =  0.1338 , as 1 - 0.78 = 0.22 is the chance of losing.

The probability of winning all 3 games can be found by (0.78)3 =  0.4745

Therefore the probability of him winning at least 2 games (the first 2 games or all games) is 0.1338 + 0.4745 = 0.608

(c)

Probability that player defeats all three opponents is 0.4745
Implies , the probability of not defeating all three opponents is 1 - 0.4745 = 0.5255
Let A denote event that player does not defeat all three opponents in all two games.
Then,
P(A) = (0.5525)2 = (0.5525)(0.5525) = 0.3052
Thus the probability that the player defeats all three opponents at least once
P(A') = 1 - P(A) = 1 - 0.3052 = 0.6948
Therefore, the probability that the player defeats all three opponents at least once is 0.6948.

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