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Electronic baseball games manufactured by Tempco Electronics are shipped in lots of \(24 .\) Before shipping, a quality-control inspector randomly selects a sample of 8 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 2 defective games will still be shipped?

Short Answer

Expert verified
The probability that a lot containing exactly 2 defective games will still be shipped is \(\frac{1}{552}\).

Step by step solution

01

Calculate the total number of ways to select 8 games from the lot

To determine the total number of ways to select 8 games from the lot of 24, we can use the combination formula, which is given by: $$C(n, r) = \frac{n!}{r!(n-r)!}$$ Here, \(n\) represents the total number of games in the lot (24) and \(r\) represents the number of games we want to choose (8). So, the total number of ways to select 8 games is: $$C(24, 8) = \frac{24!}{8!(24-8)!}$$
02

Calculate the number of ways to select 8 non-defective games

Since there are 2 defective games in the lot, there must be 22 non-defective games. We want to find the number of ways to select all 8 games from these 22 non-defective games. We can use the same combination formula: $$C(22, 8) = \frac{22!}{8!(22-8)!}$$
03

Calculate the probability

To find the probability that a sample of 8 games does not contain any defective games, we should divide the number of ways to select 8 non-defective games (Step 2) by the total number of ways to select 8 games (Step 1). $$ P(\text{no defective games}) = \frac{C(22, 8)}{C(24, 8)} = \frac{\frac{22!}{8!(22-8)!}}{\frac{24!}{8!(24-8)!}} $$
04

Simplify the expression

Now we can simplify the expression: $$ P(\text{no defective games}) = \frac{22!}{8!(22-8)!} \cdot \frac{8!(24-8)!}{24!} = \frac{22!}{14!} \cdot \frac{16!}{24!} $$ Now, we can start canceling the common terms: $$P(\text{no defective games}) = \frac{22 \cdot 21 \cdot \cdots \cdot 15}{24 \cdot 23 \cdot \cdots \cdot 17}$$
05

Calculate the probability

Multiplying and dividing the remaining terms: $$P(\text{no defective games}) = \frac{1}{23 \cdot 24} = \frac{1}{552}$$ So, the probability that a lot containing exactly 2 defective games will still be shipped is \(\frac{1}{552}\).

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Key Concepts

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

Combinatorial Analysis
Combinatorial analysis is a field of mathematics that deals with counting, enumeration, and finding the probability of specific configurations within sets. In the context of our exercise, it becomes a crucial tool in quality control processes, as it allows us to calculate the number of ways to choose a subset, in this case, a random sample of 8 games from a lot of 24.

Using the combinatorial formula known as the combination formula, we can determine the total possible selections without regard to the order of selection. This formula is written as: \[C(n, r) = \frac{n!}{r!(n-r)!}\], where \(n!\) (n factorial) is the product of all positive integers up to \(n\), and similarly for \(r!\) and \((n-r)!\). In our example, the calculation step using \(C(24, 8)\) gives us the total number of ways to select 8 games from a lot of 24.

To understand this better, consider a simpler example. If we had just 3 games and wanted to know how many ways we could choose 2, we could list the possibilities: Game 1 and Game 2, Game 1 and Game 3, or Game 2 and Game 3 — for a total of 3 combinations.
Quality Control
Quality control is a critical aspect of manufacturing and distribution that ensures products meet specific standards before reaching consumers. In our problem, Tempco Electronics uses a quality control process to inspect lots of electronic baseball games for defects. A sample of 8 games is randomly selected from each lot of 24 for testing.

The quality control process involves establishing criteria, which in this scenario, is the presence of defective games. If the sample reveals any defective games, the lot is rejected. This helps prevent the distribution of faulty products and maintains brand integrity. However, it's important to consider the implications of such a sampling method. While efficient, there's a possibility that defective items pass through if they're not included in the sample. Here, combinatorial analysis aids in evaluating these risks by quantifying the probability of selecting samples with no defects, even when some exist in the lot.

Effective quality control hinges on setting the right sample size and assessing risks accurately. This minimizes losses from rejecting good lots and prevents the shipment of defective products. Thus, combinatorial analysis is not just a mathematical tool, but also an integral part of the quality control decision-making process.
Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states and is particularly suited for scenarios involving a fixed number of trials or samples. Although our particular problem does not specifically require binomial distribution for calculation, understanding its relevance to similar quality control scenarios is beneficial.

In quality control tests where items are either defective or non-defective (two possible outcomes), and the sample is tested one unit at a time with a constant probability of finding a defective, a binomial distribution would be appropriate. The probability of finding exactly \(k\) defective items in an \(n\)-sized sample is given by the formula: \[P(X = k) = C(n, k) \cdot p^k (1-p)^{(n-k)}\], where \(p\) is the probability of finding a defective item in one trial.

However, in the given exercise, binomial distribution is not used since the sample contains games are chosen all at once, without replacement, altering the likelihood with each pick. Instead, we rely solely on combinatorial analysis to calculate the exact probability. Nevertheless, the principles of binomial distribution can still inform about the general behavior of such quality control systems when sampling is done sequentially rather than in one batch.

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