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An optical inspection system is to distinguish among different part types. The probability of a correct classification of any part is \(0.98 .\) Suppose that three parts are inspected and that the classifications are independent. Let the random variable \(X\) denote the number of parts that are correctly classified. Determine the probability mass function of \(X.\)

Short Answer

Expert verified
The PMF of X is: P(X = 0) = 0.000008, P(X = 1) = 0.001176, P(X = 2) = 0.057624, P(X = 3) = 0.941192.

Step by step solution

01

Understanding the Random Variable X

The random variable \( X \) represents the number of correctly classified parts out of three inspected parts. It can take values \( X = 0, 1, 2, \) or \( 3 \).
02

Understanding Binomial Distribution

In this scenario, each part being correctly classified is a Bernoulli trial with success probability \( p = 0.98 \). Since the inspection of three parts is independent, \( X \) follows a binomial distribution with parameters \( n = 3 \) (trials) and \( p = 0.98 \) (probability of success).
03

Binomial Probability Mass Function (PMF)

The probability mass function for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient, \( p^k \) is the probability of \( k \) successes, and \( (1-p)^{n-k} \) is the probability of \( n-k \) failures.
04

Calculating P(X = 0)

For \( X = 0 \), all parts are incorrectly classified: \[ P(X = 0) = \binom{3}{0} (0.98)^0 (0.02)^3 = 1 \times 1 \times 0.000008 = 0.000008 \]
05

Calculating P(X = 1)

For \( X = 1 \), one part is correctly classified: \[ P(X = 1) = \binom{3}{1} (0.98)^1 (0.02)^2 = 3 \times 0.98 \times 0.0004 = 0.001176 \]
06

Calculating P(X = 2)

For \( X = 2 \), two parts are correctly classified: \[ P(X = 2) = \binom{3}{2} (0.98)^2 (0.02)^1 = 3 \times 0.9604 \times 0.02 = 0.057624 \]
07

Calculating P(X = 3)

For \( X = 3 \), all three parts are correctly classified: \[ P(X = 3) = \binom{3}{3} (0.98)^3 (0.02)^0 = 1 \times 0.941192 \times 1 = 0.941192 \]
08

Summarizing the PMF

The probability mass function of \( X \) is: \[ P(X = 0) = 0.000008 \] \[ P(X = 1) = 0.001176 \] \[ P(X = 2) = 0.057624 \] \[ P(X = 3) = 0.941192 \]

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Key Concepts

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

Probability Mass Function
The probability mass function (PMF) is a key concept when working with discrete random variables like the binomial distribution. It's essentially a way of showing the probabilities of all possible outcomes for a discrete random variable. For a binomial distribution, each outcome corresponds to a certain number of successful trials out of a total number of trials.
For example, if we want to know the probability of correctly classifying exactly two parts out of three, the PMF helps us calculate this probability. The binomial PMF formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Here, \(n\) is the total number of trials, \(k\) is the number of successful trials, \(p\) is the success probability, and \(\binom{n}{k}\) is the number of ways to choose \(k\) successes from \(n\) trials.
This function helps us systematically determine the likelihood of each possible outcome, providing a complete picture of the distribution's behavior. Each value of \(k\) results in a specific probability, which together form the PMF.
Independent Trials
In many probabilistic scenarios, the concept of independent trials is crucial. An experiment is considered to have independent trials when the outcome of one trial does not affect the outcome of another. This means that the probability of success or failure remains constant throughout the process.
In our example, the inspection of each part for correct classification is an independent trial. This independence ensures that whether the first part is correctly classified does not influence the classification of the second part.
  • When trials are independent, we can multiply the probabilities of success for individual trials to get the joint probability of multiple outcomes.
  • This property is particularly important in binomial experiments, where a fixed number of independent trials are performed.
Understanding independence helps ensure accurate probability calculations and is a critical assumption for applying the binomial distribution.
Bernoulli Trials
Bernoulli trials are the building blocks of the binomial distribution. A Bernoulli trial is a random experiment where there are only two possible outcomes: success or failure.
For example, in our inspection system, each part being correctly classified (success) or not (failure) is a Bernoulli trial. The probability of success remains constant, denoted as \( p \). If the system has a success probability of 0.98 for correctly classifying a part, then failing to correctly classify has a probability of \(1 - p = 0.02\).
Bernoulli trials are foundational because they form the basis for more complex probability models, like the binomial distribution. They encapsulate the simplest form of decision-making processes where only two outcomes are considered.
  • Multiple Bernoulli trials form the structure required for a binomial distribution, where each trial is independent and follows the characteristics of a Bernoulli trial.
  • Recognizing the nature of Bernoulli trials aids in understanding how results accrue over multiple trials, leading to outcomes captured by probability mass functions.
Through Bernoulli trials, we gain a deeper insight into the mechanics behind random experiments, crucial for understanding broader statistical concepts.

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