A player of a video game is confronted with a series of 3 opponents and a (n) 77% 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 3 opponents in a game?

b. What is the probability that a player defeats at least two 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?

Question

Mathematics

Jaslyn Beck

4 September, 05:49

A player of a video game is confronted with a series of 3 opponents and a (n) 77% 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 3 opponents in a game?

b. What is the probability that a player defeats at least two 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?

+3

Answers (1)

Know the Answer?

Rene Logan · Answer 1

a.) 0.4565

b.) 0.8656

c.) 0.4615

Step-by-step explanation:

We solve this using the probability distribution formula of combination.

nCr * p^r * q^n-r

Where

n = number of trials

r = successful trials

probability of success = p = 77% = 0.77

Probability of failure = q = 1-0.77 = 0.23

a.) When exactly 3 opponents are defeated, When n = 3 and r = 3, probability becomes:

= 3C3 * 0.77³ * 0.23^0

= 1 * 0.456533 * 1

= 0.456533 = 0.4565 (4. d. p)

b.) When at least 2 opponents are defeated, that is when r = 2 and when r = 3,

When r = 2, probability becomes:

= 3C2 * 0.77² * 0.23¹

= 3 * 0.5929 * 0.23

= 0.409101

When 3 opponents are defeated, we calculated it earlier to be 0.456533

Hence, probability that at least 2 opponents are defeated

= 0.409101 + 0.456533

= 0.865634 = 0.8656 (2. d. p)

c.) If 2 games are played, probability he defeat all 3 at least once in the game will be the sum (probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game) + (probability of defeating all three opponents in both games)

Probability of defeating all three opponents in the first game = 0.456533

Probability of not defeating all three opponents in the second game = 1 - 0.456533 = 0.543467

Hence,

probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game = 0.465633 * 0.543467 = 0.253056

probability of defeating all three opponents in both games

= 0.456533 * 0.456533

=0.208422

Probability he defeats all three opponents at least once in 2games

= 0.253056 + 0.208422

=0.461478 = 0.4615 (4. d. p)