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A player of a video game is confronted with a series of four opponents and an \(80 \%\) 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 four 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 three times, what is the probability that the player defeats all four opponents at least once?

Short Answer

Expert verified
(a) 0.4096; (b) 0.64; (c) 0.7939.

Step by step solution

01

Probability of Defeating One Opponent

The probability of defeating one opponent is given as 80%, or 0.8. This will be used as the base probability for defeating each opponent.
02

Probability of Defeating All Four Opponents

Each opponent is independent, so the probability of defeating all four opponents is the product of individual probabilities: \( P(\text{Defeating all 4}) = 0.8 \times 0.8 \times 0.8 \times 0.8 \). Calculating this gives: \( P(\text{Defeating all 4}) = 0.8^4 = 0.4096 \).
03

Determine Probability of Defeating Exactly 0 or 1 Opponent

First calculate the probability of defeating 0 opponents: \( P(\text{Defeating 0}) = 0.2^1 = 0.2 \) (only 1 scenario, lose to the first opponent). Then calculate for exactly 1 opponent: \( P(\text{Defeating 1}) = 0.8 \times 0.2 = 0.16 \) (win against first, lose to second).
04

Calculate Probability of Defeating At Least Two Opponents

Use the complement rule: \( P(\text{At least 2}) = 1 - (P(\text{Defeating 0}) + P(\text{Defeating 1})) \). From Step 3, \( P(\text{At least 2}) = 1 - (0.2 + 0.16) = 1 - 0.36 = 0.64 \).
05

Probability of Defeating All Opponents in Single Game

From Step 2, the probability of defeating all four opponents in a single game is \( 0.4096 \). This probability will be used to calculate further scenarios.
06

Probability of Winning At Least Once in Three Games

Calculate the probability of NOT defeating all four opponents in a single game, which is \( 1 - 0.4096 = 0.5904 \). The probability of not defeating all four opponents in any of the three games is \( 0.5904^3 \). Therefore, the probability of defeating all four at least once is \( 1 - 0.5904^3 = 1 - 0.2061 = 0.7939 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
Understanding independent events is crucial in probability, especially in scenarios like the current exercise. Independent events are two or more events that do not influence each other's outcomes. In other words, the outcome of one event doesn't affect the probability of another event occurring.

In the context of the exercise, the player has an 80% chance of defeating each opponent. The result against each opponent is independent. This means that defeating or being defeated by one opponent doesn’t change the chances against another.

Therefore, the probability of defeating all four opponents is found by multiplying the probability of defeating each opponent. Since these events are independent, the calculation is straightforward:
  • Defeat First: 0.8
  • Defeat Second: 0.8
  • Defeat Third: 0.8
  • Defeat Fourth: 0.8
By multiplying these probabilities, you get: \[ P(\text{All 4}) = 0.8 \times 0.8 \times 0.8 \times 0.8 = 0.8^4 = 0.4096 \] Knowing each event is independent helps tailor more accurate probability predictions in games and real-life scenarios.
Complement Rule
The Complement Rule is a useful tool in probability calculations. It's used to find the probability of an event by subtracting the probability of its complement from one. Basically, if you know the probability of not doing something, you can easily find the probability of doing it by using this rule.

The exercise demonstrates this with the probability of defeating at least two opponents. Instead of calculating the probability of defeating two, three, or four opponents directly, the complement rule simplifies it. You first determine the probabilities of scenarios we do not want:
  • Defeating 0 opponents: \( 0.2 \) (lose immediately)
  • Defeating exactly 1 opponent: \( 0.16 \) (win one, then lose)
The sum of these probabilities is \( 0.36 \), representing the probability of defeating fewer than two opponents.

Apply the complement rule:\[ P(\text{At least two wins}) = 1 - (0.2 + 0.16) = 1 - 0.36 = 0.64 \]This rule is especially beneficial when dealing with probabilities involving multiple potential outcomes, as shown here with fewer steps and less calculation complexity.
Binomial Probability
Binomial probability is often employed in situations featuring a fixed number of trials, each with two possible outcomes such as success or failure. Here, defeating an opponent is a success, and each game against an opponent is an independent trial.

The exercise transforms into a binomial experiment when you assess the probability of defeating a certain number of opponents out of a total, for instance, at least two. A binomial probability formula plays a role when evaluating these scenarios comprehensively:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where:
  • \( n \) is the total number of trials (or opponents).
  • \( k \) is the number of successes we're interested in.
  • \( p \) is the probability of winning against one opponent.
Even if not explicitly needed to solve this problem step-by-step, knowing binomial probability helps understand the architecture of probability in such scenarios. By connecting independent events with multiple trials, mathematicians and statisticians can predict outcomes accurately in complex sequences.
Game Theory in Probability
Game theory explores strategic interactions where outcomes depend not just on one's actions but also on others' actions. It is a field that combines probability theory for decision making and analyzing possible outcomes in games and real-life situations.

In the video game problem, the player faces opponents with a known probability of success. This sets the scene for strategic planning, leaning on game theory principles where the player can evaluate the risk (probabilities) of continuing in the game against possible rewards (winning against more opponents).

While the exercise purely explores probabilities, integrating game theory could involve considering different strategies such as:
  • Quit after defeating a certain number of opponents to prevent potential loss.
  • Asses past game outcomes to adapt the approach.
The frequent use of probabilities allows players and strategists to design optimal paths within games. Game theory, when combined with probability metrics, offers a broader understanding and tools to enhance decision making in situations with uncertainties and competitive elements.

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