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An optical inspection system is used 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: \(P(X=0)\approx0.000008, P(X=1)\approx0.000024, P(X=2)\approx0.057648, P(X=3)\approx0.941192\).

Step by step solution

01

Define the Random Variables

We have three parts, each of which can be either correctly classified or not. Define the random variable for each part being correctly classified. Let each part being correctly classified be represented by a Bernoulli trial with probability of success (correct classification) as 0.98.
02

Identify the Number of Trials

There are three trials here because we are inspecting three parts. Hence, each part classification is a trial, resulting in three independent Bernoulli trials.
03

Determine Probability of Success and Failure

The probability of success, which is correctly classifying a part, is 0.98. Consequently, the probability of failure (incorrect classification) is 1 - 0.98 = 0.02.
04

Use Binomial Distribution

Since the classifications are independent and each part can be classified as correct or incorrect, the total number of correctly classified parts follows a Binomial distribution with parameters: number of trials \(n = 3\) and probability of success \(p = 0.98\).
05

Write the Probability Mass Function

The probability mass function for a Binomial distribution is given by \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \(n = 3\), \(p = 0.98\), and \(k\) ranges from 0 to 3.
06

Calculate Probability Values

Compute the probabilities for \(k = 0, 1, 2, 3\):- For \(k = 0\): \[ P(X = 0) = \binom{3}{0} (0.98)^0 (0.02)^3 = 1 \times 0.02^3 \approx 0.000008 \]- For \(k = 1\):\[ P(X = 1) = \binom{3}{1} (0.98)^1 (0.02)^2 = 3 \times 0.98 \times 0.0004 \approx 0.000024 \]- For \(k = 2\):\[ P(X = 2) = \binom{3}{2} (0.98)^2 (0.02)^1 = 3 \times 0.9604 \times 0.02 \approx 0.057648 \]- For \(k = 3\):\[ P(X = 3) = \binom{3}{3} (0.98)^3 (0.02)^0 = 1 \times 0.98^3 \approx 0.941192 \]
07

Summarize the Probability Mass Function

The probability mass function of \(X\) is given by:- \(P(X = 0) \approx 0.000008\)- \(P(X = 1) \approx 0.000024\)- \(P(X = 2) \approx 0.057648\)- \(P(X = 3) \approx 0.941192\)

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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 crucial in understanding how probabilities are distributed across different numbers of successful outcomes in a discrete random process. In our case, a PMF describes how likely it is to correctly classify 0, 1, 2, or 3 parts out of 3 inspected parts using an optical system.
The PMF is defined for discrete random variables, which means it gives probabilities for specific outcomes rather than ranges, as continuous distributions do.
  • It provides a complete description of the probability distribution for a Binomial random variable.
  • The PMF for a binomial distribution is calculated as \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(n\) is the number of trials, \(p\) is the probability of success, and \(k\) is the number of successful trials.
This helps us understand the likelihood of correctly classifying a specific number of parts and can be crucial for quality control and predicting performance in manufacturing systems.
Bernoulli Trials
Bernoulli trials are the basic building blocks for many probability distributions, including the binomial distribution. Each trial here refers to the classification of one part, with two possible outcomes: correctly classified or not.
Each Bernoulli trial is characterized by:
  • A binary outcome (Success or Failure).
  • A constant probability of success, denoted by \(p\), which in this exercise is 0.98.
These trials are crucial because they form the basis of the Binomial distribution, applied when we analyze the probability of correctly classifying a certain number of parts. When multiple Bernoulli trials are conducted, as with the three parts we inspect here, we consider them collectively under a Binomial distribution if each trial is independent and the probability of success remains constant across trials.
Probability of Success
The probability of success \(p\) is central to calculating binomial probabilities. In this context, it is the probability that a part is correctly classified by the system, which is given as 0.98.
Understanding \(p\) helps determine not only individual probabilities but also overall system performance.
  • High values of \(p\) indicate that the system is quite reliable in its classifications.
  • The complement, or probability of failure, is \(1 - p\). Here, it equals 0.02, highlighting a low chance of misclassification.
By maximizing \(p\) and minimizing the probability of failure, systems can ensure greater accuracy and reliability, especially in industries where precision is crucial. These values are integral in shaping expectations and evaluating system effectiveness.
Independent Events
Independence in probability means the outcome of one event does not influence or change the likelihood of another. When parts are independently classified, as in this exercise, it implies that classifying one part correctly does not affect the chances of classifying another correctly.
The importance of independent events lies in simplifying calculations and assumptions:
  • For independent events, the total probability of compound events can be found by multiplying the individual probabilities.
  • This principle is central to the binomial model since the independence allows the use of the specific binomial formula without additional adjustments for dependency.
This independence assures that each trial is unaffected by the outcomes of others, enabling straightforward use of the binomial distribution formula to calculate exact probabilities across multiple trials.

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Most popular questions from this chapter

A company performs inspection on shipments from suppliers to detect nonconforming products. Assume that a lot contains 1000 items and \(1 \%\) are nonconforming. What sample size is needed so that the probability of choosing at least one nonconforming item in the sample is at least \(0.90 ?\) Assume that the binomial approximation to the hypergeometric distribution is adequate.

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