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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 contest opponents. a. What is the probability mass function of the number of opponents contested in a game? b. What is the probability that a player defeats at least two opponents in a game? c. What is the expected number of opponents contested in a game? d. What is the probability that a player contests four or more opponents in a game? e. What is the expected number of game plays until a player contests four or more opponents?

Short Answer

Expert verified
a. PMF is \( P(X=k) = 0.20^{k-1} \cdot 0.80 \). b. Probability is 0.04. c. Expected opponents is 1.25. d. Probability is 0.008. e. Expected plays is 125.

Step by step solution

01

Identify the Distribution

This scenario can be described using the geometric distribution. The geometric distribution models the number of trials until the first failure occurs in a sequence of independent Bernoulli trials. In this situation, the trials represent contests against opponents, with a success (defeat) probability of 0.80.
02

Probability Mass Function (PMF)

The probability mass function of a geometric distribution is given by \( P(X=k) = (1-p)^{k-1} \cdot p \), where \( p \) is the probability of success on each trial. For this problem, \( p = 0.80 \). Thus, the PMF is \( P(X=k) = 0.20^{k-1} \cdot 0.80 \).
03

Probability of Defeating At Least Two Opponents

We want to find \( P(X \geq 2) \), which can be calculated as 1 minus the probability of defeating no opponents or exactly one opponent. Using the PMF: \( P(X=1) = 0.80 \) and \( P(X=2) = 0.20 \cdot 0.80 = 0.16 \). Therefore, \( P(X \geq 2) = 1 - (0.80 + 0.16) = 0.04 \).
04

Expected Number of Opponent Contests

The expected value for the number of trials in a geometric distribution is \( E(X) = \frac{1}{p} \). Substituting our \( p = 0.80 \), \( E(X) = \frac{1}{0.80} = 1.25 \).
05

Probability of Contesting Four or More Opponents

We need to find \( P(X \geq 4) \). This is equal to 1 minus the probability of contesting fewer than 4 opponents. Calculate \( P(X=1) = 0.80 \), \( P(X=2) = 0.16 \), and \( P(X=3) = 0.20^2 \cdot 0.80 = 0.032 \). Therefore, \( P(X \geq 4) = 1 - (0.80 + 0.16 + 0.032) = 0.008 \).
06

Expected Number of Plays for Contesting 4 or More

The mean or expected value in a geometric distribution for success is \( \frac{1}{P(X \geq 4)} \). From Step 5, \( P(X \geq 4) = 0.008 \). Thus, the expected number of games is \( \frac{1}{0.008} = 125 \).

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Key Concepts

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

Probability Mass Function
In the context of the geometric distribution, the probability mass function (PMF) is a tool to determine the likelihood of a specific number of trials before obtaining the first failure. For the video game scenario, every time the player faces an opponent, it's considered a Bernoulli trial, a trial which only has two outcomes: success (defeating the opponent) or failure.

The PMF of a geometric distribution is expressed as: \[ P(X=k) = (1-p)^{k-1} \cdot p \] Here, \(p\) is the probability of success (defeating an opponent), which is 0.80. The formula shows that the player is more likely to face fewer opponents before a loss, but the exact probability for facing \(k\) opponents varies based on \(k\).

For instance, if you want to find the probability that the player faces exactly 3 opponents, you simply substitute \(k=3\) into the PMF formula.
Expected Value
The expected value is a key concept in probability that provides the average number of trials we can anticipate before experiencing the first failure. In a geometric distribution, the expected value is inversely proportional to the probability of success. That means, \[ E(X) = \frac{1}{p} \]

For the video game problem, each fight with an opponent represents a trial. With a success probability of 0.80, the expected number of opponents the player can defeat before losing is \[ E(X) = \frac{1}{0.80} = 1.25 \] This suggests that, on average, the player can defeat about one to two opponents before getting defeated.

The expected value is useful because it gives us a sense of the behavior over many games, even if in any single game the player might defeat more or fewer opponents.
Bernoulli Trials
Bernoulli trials are fundamental in probability and statistics, forming the basis for many types of distributions, including the geometric distribution. Each Bernoulli trial has exactly two possible outcomes: success or failure. For the video game, confronting each opponent is a Bernoulli trial where the player either wins (success) or loses (failure).

The trials are independent, meaning the outcome of facing one opponent doesn't affect the outcomes with subsequent opponents. This independence is crucial for the validity of the geometric distribution. Each time the player battles an opponent, there's a consistent 80% chance of winning, regardless of prior results.

Understanding Bernoulli trials helps to comprehend complex statistical concepts and models that build upon this simplicity.
Probability Calculations
Probability calculations with the geometric distribution often involve determining the chances of achieving a particular outcome after a sequence of trials. This includes calculating the probability of certain numbers of successes or failures before a failure. For example, the probability that the player contests at least two opponents is found by evaluating \[ P(X \geq 2) = 1 - (P(X=1) + P(X=2)) \]

Using the PMF, we know \(P(X=1) = 0.80\) and \(P(X=2) = 0.16\), so \[ P(X \geq 2) = 1 - (0.80 + 0.16) = 0.04 \]

Another essential calculation might be determining the probability of contesting four or more opponents: \[ P(X \geq 4) = 1 - (P(X=1) + P(X=2) + P(X=3)) = 0.008 \]

By mastering these calculations, you can effectively predict the likelihood of various outcomes in scenarios modeled by geometric distributions.

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