Problem 152 - Applied Statistics and Probability for Engineers

Question

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?

Step-by-Step Solution

Answer

(a) 0.4096; (b) 0.5952; (c) 0.7946.

Step 1: Determine probability of defeating all opponents

The player has an 80% chance of defeating each opponent. The probability of defeating all four opponents is the product of the probability of defeating each one, since the events are independent:\[P(A) = 0.8 \times 0.8 \times 0.8 \times 0.8 = (0.8)^4 = 0.4096\]

Step 2: Calculate the probability of defeating fewer than two opponents

To find the probability of defeating at least two opponents, first calculate the probability of defeating 0 or 1 opponent and subtract it from 1.- Probability of defeating 0 opponents (loses the first one): \[ P(0) = (1 - 0.8) = 0.2 \]- Probability of defeating exactly 1 opponent: Defeats first and loses the second, third, or fourth. \[ P(1) = 0.8 \times 0.2 = 0.8 \times 0.2^3 = 0.128\] But we have 4 scenarios for exactly one win (Wins 1st, 2nd, 3rd, or 4th): \[ P(1) = 4 \times 0.2 \times (0.8) \times (0.2)^2 = 0.2048 \]Now subtract these from 1:\[P(\text{at least 2}) = 1 - (P(0) + P(1)) = 1 - (0.2 + 0.2048) = 0.5952\]

Step 3: Calculate Probabilty of Defeating All Opponents At Least Once in 3 Games

To find the probability that the player defeats all four opponents at least once in three games, calculate the probability of not defeating all four opponents in any of the games and subtract from 1.- Probability of not defeating all four opponents in a single game: \[ P( ext{not A}) = 1 - P(A) = 1 - 0.4096 = 0.5904 \]- Probability of not defeating all four opponents in three games: \[P( ext{not A in all 3}) = (0.5904)^3 = 0.2054\]- Probability of defeating all four at least once: \[P( ext{at least one A}) = 1 - 0.2054 = 0.7946\]

Key Concepts

Understanding Independent Events

When we say events are independent in probability theory, it means the outcome of one event does not affect the outcome of another. For example, in the context of the video game, defeating the first opponent has no impact on the probability of defeating the second, third, or fourth opponent. This occurrence of independent events simplifies calculations because the overall probability of a sequence of independent events happening is the product of their individual probabilities.
In mathematical terms, if events A and B are independent, then:

• \( P(A \text{ and } B) = P(A) \times P(B) \)

In our exercise, since each opponent encounter is independent, we can calculate the probability of defeating all opponents by multiplying the probability of defeating each one. Consequently, the probability that the player defeats all four opponents is \( (0.8)^4 \). Independent events often simplify probability problems, making them easier and less complex to solve.

Exploring Binomial Probability

Binomial probability deals with experiments or trials where there are exactly two possible outcomes. In our scenario, the player either defeats an opponent or doesn't. This type of situation is perfectly modeled by a binomial distribution, which is essentially a series of Bernoulli trials (independent binary outcomes).
The formula for binomial probability is:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

Where:

• \( n \) is the number of trials (opponents faced).
• \( k \) is the number of successful outcomes (opponents defeated).
• \( p \) is the probability of success on each trial (defeating an opponent).

Calculating the probability of defeating at least two opponents involves understanding all combinations where the player wins 2, 3, or all 4 rounds, and the simplest way to find this aggregate probability in this case was to subtract from 1 the probability of defeating 0 or 1 opponent only.

Decoding Conditional Probability

Conditional probability helps analyze the likelihood of an event occurring, given that another event has already occurred. In some cases, events might not be independent, and the occurrence of one event could affect the probability of another. However, in our problem, events are independent, yet understanding conditional probability can be insightful as it applies more broadly in probability theory.
Conditional probability is formalized as:

• \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

Where \( P(A|B) \) is the probability of event A happening given that B has happened. In our specific exercise, while these concepts aren't directly used due to the independence of events, it's valuable in scenarios where the result of one event influences another.
For example, if the probability of winning three games in a row mattered after losing one, conditional probability would be essential in recalculating those odds. This highlights the diverse application of conditional probability beyond independent events and shows its importance in broader probability calculations.