Question

A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to context opponents. 1). What is the probability that a player defeats at least three opponents in a game

Answer 1

0.512 = 51.2% probability that a player defeats at least three opponents in a game

Explanation:

Outcomes in which he defeats less than three opponents:

Lose to the first opponent: 0.2 probability.

Beats the first opponent, loses to the second: 0.8*0.2 probability

Beats the first two opponents, loses to the third: 0.8*0.8*0.2 probability:

Probability of losing before beating three opponents:

`P(b < 3) = 0.2 + 0.8*0.2 + 0.8*0.8*0.2 = 0.488`

What is the probability that a player defeats at least three opponents in a game?

`P(b \geq 3) = 1 - p(b < 3) = 1 - 0.488 = 0.512`

0.512 = 51.2% probability that a player defeats at least three opponents in a game