91Ó°ÊÓ

Four electronic ovens that were dropped during shipment are inspected and classified as containing either a major, a minor, or no defect. In the past, \(60 \%\) of dropped ovens had a major defect, \(30 \%\) had a minor defect, and \(10 \%\) had no defect. Assume that the defects on the four ovens occur independently. (a) Is the probability distribution of the count of ovens in each category multinomial? Why or why not? (b) What is the probability that, of the four dropped ovens, two have a major defect and two have a minor defect? (c) What is the probability that no oven has a defect? Determine the following: (d) The joint probability mass function of the number of ovens with a major defect and the number with a minor defect (e) The expected number of ovens with a major defect (f) The expected number of ovens with a minor defect (g) The conditional probability that two ovens have major defects given that two ovens have minor defects (h) The conditional probability that three ovens have major defects given that two ovens have minor defects (i) The conditional probability distribution of the number of ovens with major defects given that two ovens have minor defects (j) The conditional mean of the number of ovens with major defects given that two ovens have minor defects.

Step by step solution

01

Determine if distribution is multinomial

The problem describes a scenario where we examine four electronic ovens that can each be classified into one of three categories: major defect, minor defect, or no defect. The characteristics are independent and each oven falls into only one category with fixed probabilities: 60% for major defect, 30% for minor defect, and 10% for no defect. This aligns with the definition of a multinomial distribution, which describes the probabilities of counts across multiple categories based on certain probabilities and a fixed number of trials. Thus, the distribution is multinomial.

02

Probability of two major and two minor defects

Using the multinomial probability formula \[P(X_1 = x_1, X_2 = x_2, ..., X_k = x_k) = \frac{n!}{x_1! x_2! ... x_k!} p_1^{x_1} p_2^{x_2} ... p_k^{x_k}\] where \(n\) is the total number of trials, and \(p_i\) is the probability of choosing category \(i\). For two major defects (\(x_1 = 2\) with \(p_1 = 0.6\)) and two minor defects (\(x_2 = 2\) with \(p_2 = 0.3\)), we find:\[P = \frac{4!}{2!2!0!} (0.6)^2 (0.3)^2 = 6 \times 0.36 \times 0.09 = 0.1944\].

03

Probability no oven has a defect

To find the probability that all four ovens have no defects, we use \(p_3 = 0.1\) for four ovens falling into the no defect category. Using the probability formula: \[P(X_3 = 4) = \frac{4!}{0!0!4!} (0.1)^4 = 1 \times (0.1)^4 = 0.0001\].

04

Determine joint probability mass function

The joint probability mass function is determined for two random variables: the number of major defects \(M\), and the number of minor defects \(m\), defined as \[P(M = m, N = n) = \frac{n!}{m!n!(n-m)!}(0.6)^m(0.3)^n(0.1)^{n-m}\].

05

Expected number of major defects

The expected value for major defects is calculated using \(E(M) = n \cdot p_{\text{major}}\). With three ovens, \(n = 4\) and \(p_{\text{major}} = 0.6\), giving \[E(M) = 4 \times 0.6 = 2.4\].

06

Expected number of minor defects

The expected number for minor defects is given by \(E(m) = n \cdot p_{\text{minor}}\). For four ovens, \(n = 4\) and \(p_{\text{minor}} = 0.3\), resulting in \[E(m) = 4 \times 0.3 = 1.2\].

07

Conditional probability of two major defects given two minor

The probability for exactly two major and two minor defects is calculated as in Step 2, which is 0.1944. The probability of exactly two minor defects independently is given by \[P(m = 2) = \frac{4!}{2!2!0!}(0.3)^2(0.1)^2(0.6)^0 = 0.054\]. \[P(2\ \text{major} \mid 2\ \text{minor}) = \frac{0.1944}{0.054} \approx 3.6\].

08

Conditional probability of three major defects given two minor

Calculate the probability of exactly three major defects and two minor defects. Here, as these two cannot occur due to sum constraint, \(P(3\ \text{major}, 2\ \text{minor}) = 0\). So, \[P(3\ \text{major} \mid 2\ \text{minor}) = 0\].

09

Conditional distribution of number of major defects given two minor

Considering the distribution given two minor defects, the potential possibilities for major defects are 0, 1, and 2. Calculate respective probabilities, noting impossible configurations do not contribute (as checked in previous steps). Thus, \(P(M = x \mid m = 2)\) is only valid for feasibly computed entries.

10

Conditional mean of number of major defects

The conditional mean is obtained by summing the product of probabilities and corresponding major defects given two minor defects: \[E(M \mid m = 2) = \sum_{x_i} x_i \cdot P(M = x_i \mid m = 2)\], with contributions from non-zero values above (distribution found in Step 9).

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Key Concepts

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

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood of events happening. It provides a formal framework for quantifying uncertainty through probabilities, which range from 0 (impossible event) to 1 (certain event). In the context of the multinomial distribution, probabilities are used to model scenarios where outcomes fall into multiple categories. Understanding this is key for solving problems involving defect distributions among ovens.
For example, if we drop ovens and categorize them based on defects, we can assign fixed probabilities to each category: major, minor, and no defect. The sum of these probabilities is always 1, reflecting comprehensive categorization. Probability theory then helps in calculating specific likelihoods, like finding the probability of a certain number of ovens having major or minor defects.

Statistical Independence

In probability theory, two events are statistically independent if the occurrence of one does not affect the probability of the other happening. This is crucial when working with distributions like the multinomial distribution. In the oven example, each oven's defect status is said to be statistically independent, meaning the defect probability of one oven does not affect another.
Independence can be determined by checking if the probability of both events occurring together equals the product of their individual probabilities. In practical terms, achieving statistical independence simplifies calculations, allowing us to multiply probabilities without adjustment for interactions between events.

Expected Value

The expected value in probability theory is the average outcome you'd expect if an experiment were repeated many times under the same conditions. It's a crucial concept used to quantify the central tendency of a random variable.
In the exercise involving defective ovens, the expected number of major defects is computed using the known probability of each category. Specifically, for the major defects, the expected value, denoted as \(E(M)\), is calculated by multiplying the total number of trials (ovens) by the probability of the major defect category. Thus, \[E(M) = n \times p_{\text{major}}\], which offers insight into typical outcomes when dealing with similar scenarios over time.

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already occurred. It's a fundamental concept that refines our understanding when information is available about related events. Conditional probability is denoted by \(P(A \mid B)\), representing the probability of event A occurring given that B is true.
In the context of the exercise, we consider conditional probabilities like the probability of having two major defects given two minor defects. This requires dividing the joint probability of both conditions by the probability of the condition that's given (in this case, two minor defects). Formally, this is expressed as \[P(2 \ \text{major} \mid 2 \ \text{minor}) = \frac{P(2 \ \text{major}, 2 \ \text{minor})}{P(2 \ \text{minor})}\]. This helps refine probability calculations based on prior known outcomes.