/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 A lot consists of \(N\) transist... [FREE SOLUTION] | 91Ó°ÊÓ

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A lot consists of \(N\) transistors, and of these, \(M\) \((M \leq N)\) are defective. We randomly select two transistors without replacement from this lot and determine whether they are defective or nondefective. The random variable $$X_{i}=\left\\{\begin{array}{ll}1, & \text { if the } i \text { th transistor } \\\& \text { is nondefective } \\\0, & \text { if the } i \text { th transistor } \\\& \text { is defective }\end{array} i=1,2\right.$$ Determine the joint probability function for \(X_{1}\) and \(X_{2}\). What are the marginal probability functions for \(X_{1}\) and \(X_{2}\) ? Are \(X_{1}\) and \(X_{2}\) independent random variables?

Short Answer

Expert verified
The joint probability function is given by calculated probabilities of each outcome; the marginals are \( P(X_1=0) = \frac{M}{N}, P(X_1=1) = \frac{N-M}{N}, P(X_2=0) = \frac{M}{N}, P(X_2=1) = \frac{N-M}{N} \); \(X_1\) and \(X_2\) are not independent.

Step by step solution

01

Understand the Random Variables

The random variables \(X_{1}\) and \(X_{2}\) represent whether the first and second transistors chosen are nondefective (value of 1) or defective (value of 0). We need to determine the probability mass function for these variables.
02

Determine Possible Outcomes

The possible outcomes for \( (X_{1}, X_{2}) \) are: \((0,0)\), both defective; \((1,0)\), the first nondefective, second defective; \((0,1)\), the first defective, second nondefective; and \((1,1)\), both nondefective.
03

Calculate Joint Probability for (0,0)

First transistor defective: \( \frac{M}{N} \). Second transistor defective given first is defective: \( \frac{M-1}{N-1} \). Thus, \( P(X_{1}=0, X_{2}=0) = \frac{M}{N} \cdot \frac{M-1}{N-1} \).
04

Calculate Joint Probability for (1,0)

First transistor nondefective: \( \frac{N-M}{N} \). Second transistor defective given first is nondefective: \( \frac{M}{N-1} \). Thus, \( P(X_{1}=1, X_{2}=0) = \frac{N-M}{N} \cdot \frac{M}{N-1} \).
05

Calculate Joint Probability for (0,1)

First transistor defective: \( \frac{M}{N} \). Second transistor nondefective given first is defective: \( \frac{N-M}{N-1} \). Thus, \( P(X_{1}=0, X_{2}=1) = \frac{M}{N} \cdot \frac{N-M}{N-1} \).
06

Calculate Joint Probability for (1,1)

First transistor nondefective: \( \frac{N-M}{N} \). Second transistor nondefective given first is nondefective: \( \frac{N-M-1}{N-1} \). Thus, \( P(X_{1}=1, X_{2}=1) = \frac{N-M}{N} \cdot \frac{N-M-1}{N-1} \).
07

Marginal Probability Function for X1

\( P(X_{1}=0) = P(X_{1}=0, X_{2}=0) + P(X_{1}=0, X_{2}=1) = \frac{M}{N} \). \( P(X_{1}=1) = P(X_{1}=1, X_{2}=0) + P(X_{1}=1, X_{2}=1) = \frac{N-M}{N} \).
08

Marginal Probability Function for X2

\( P(X_{2}=0) = P(X_{1}=0, X_{2}=0) + P(X_{1}=1, X_{2}=0) = \frac{M}{N} \). \( P(X_{2}=1) = P(X_{1}=0, X_{2}=1) + P(X_{1}=1, X_{2}=1) = \frac{N-M}{N} \).
09

Check Independence of X1 and X2

\(X_{1}\) and \(X_{2}\) are independent if \( P(X_{1}=i, X_{2}=j) = P(X_{1}=i)P(X_{2}=j) \) for all \(i, j\). Here, due to the without replacement context, \(X_{1}\) and \(X_{2}\) are not independent.

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Key Concepts

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

Marginal Probability
Marginal probability offers a way to analyze individual probabilities separate from the joint outcomes. It allows us to find the likelihood of a single event occurring, regardless of the outcome of another variable. This is particularly useful for complex problems where multiple variables interact.

In the context of our transistors problem, we calculated the marginal probability for each random variable separately. Let's take for instance, random variable \(X_1\). It represents whether the first selected transistor is nondefective or defective:
  • \(P(X_1 = 0)\), the probability that the first transistor is defective, can be found by summing up the probabilities of \((0,0)\) and \((0,1)\).
  • Similarly, \(P(X_1 = 1)\), the probability that the first transistor is nondefective, is calculated by summing the probabilities of \((1,0)\) and \((1,1)\).
For both random variables \(X_1\) and \(X_2\), these probabilities might vary in different scenarios, greatly influencing decision-making and prediction analysis in real-world applications.
Random Variables
Random variables are core components of probability theory, representing numerical outcomes of random phenomena. They allow us to predict probabilities associated with various outcomes. In this case, \(X_1\) and \(X_2\) are random variables corresponding to the condition (defective or nondefective) of the two selected transistors.
  • Each random variable can take on specific values, in our case 0 or 1, representing defective and nondefective respectively.
  • The joint distribution of these random variables helps to understand the relationship between them.
  • It's important to note that the practice of modeling real-world problems often involves defining and analyzing such random variables to simplify the understanding of probabilistic events.
Understanding random variables and how they relate to each other is crucial for solving problems involving chance and uncertainty. This understanding sets the groundwork for further statistical analyses and predictions.
Independence of Events
In probability theory, two events are considered independent if the occurrence of one does not affect the probability of the other. However, in scenarios where items are selected without replacement, such as in our transistors problem, the events are usually dependent.

In our exercise, the transistors are drawn without replacement, meaning once one is picked, it cannot be chosen again. This dependency affects the independence between \(X_1\) and \(X_2\):
  • If \(X_1\) is defective, it changes the probability distribution for \(X_2\), either reducing the number of potentially defective transistors or nondefective ones.
  • This dependency can be tested through checking if \(P(X_1 = i, X_2 = j) = P(X_1 = i)P(X_2 = j)\) for all possible outcomes \(i, j\). The failure of this condition indicates dependence, as observed in our problem.
Understanding whether events are independent is vital. It helps in constructing accurate models and simulations in probability and statistics, influencing predictions and outcomes in various fields such as engineering, economics, and data science.

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