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Chapter 2: Problem 49

A bin of 50 parts contains 5 that are defective. A sample of 10 parts is selected at random, without replacement. How many samples contain at least four defective parts?

Short Answer

Expert verified
There are 4,292 samples containing at least four defective parts.

Step by step solution

01

Understand the Total Number of Possible Samples

The problem asks for the number of ways to choose a sample of 10 from 50. This is a combinations problem because the order of selection doesn't matter.We use the combination formula:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \(n\) is the total number, and \(k\) is the number of selections.Applying it here, we find:\[ \binom{50}{10} = \frac{50!}{10! \times 40!} \]
02

Calculate the Number of Ways to Select 4 Defective Parts

We need to find how many ways we can select exactly 4 defective parts out of the 5 defective parts available.Use the combination formula:\[ \binom{5}{4} = \frac{5!}{4! \times (5-4)!} = 5 \]
03

Calculate the Number of Ways to Select 6 Non-defective Parts

After selecting 4 defective parts, we need to select the remaining 6 parts from the non-defective ones.There are \(50 - 5 = 45\) non-defective parts. Use the combination formula:\[ \binom{45}{6} = \frac{45!}{6! \times 39!} \]
04

Calculate Samples with Exactly 5 Defective Parts

Now, calculate the number of samples containing exactly 5 defective parts.Select 5 defective parts:\[ \binom{5}{5} = 1 \]Select 5 non-defective parts:\[ \binom{45}{5} = \frac{45!}{5! \times 40!} \]
05

Add Results to Find Total Samples with At Least 4 Defective Parts

Now, add the results for both 4 defective and 5 defective cases.Calculate the total for exactly 4 defective parts:\[ \binom{5}{4} \times \binom{45}{6} \]Add this to the result for exactly 5 defective parts:\[ \binom{5}{5} \times \binom{45}{5} \]Calculate the total: Sum the results of both calculations.

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Key Concepts

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

Understanding Combinations
In combinatorial analysis, the concept of combinations is crucial. Combinations allow us to determine the number of ways to select items from a group where order doesn't matter. For this exercise, we're asked to choose parts out of a bin. To calculate this, we use the combination formula:
  • \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
  • "n" is the total number of items
  • "k" is the number of selections.
For example, choosing 10 parts from 50 would look like \( \binom{50}{10} \). This formula helps us simplify large numbers of selections into understandable calculations.
The Concept of Sampling Without Replacement
Sampling without replacement happens when we select items from a set and do not return them. This means that each selection reduces the total pool to choose from. In our exercise, we selected 10 parts without replacing them, impacting our subsequent choices.
When you pick without replacement, each choice alters future probabilities, as each item picked alters the composition of the remaining set. This idea is significant in scenarios like quality control testing, where each sampled item provides information about the batch.
Defective and Non-Defective Parts
Every part in our bin can either be defective or non-defective. Understanding this distinction helps in determining successful outcomes or events. In our problem, we have 50 parts with 5 being defective. We are interested in samples with at least four defective parts.
  • Calculate based on defective samples: Determine how many ways to select the defective parts, using \(\binom{5}{4}\) or \(\binom{5}{5}\).
  • Complement with non-defective samples: For each scenario of selecting defects, calculate corresponding non-defective selections, such as \(\binom{45}{6}\) or \(\binom{45}{5}\).
This logical split allows for easy tracking of possible desired outcomes.
Probability Calculation in Combinatorial Problems
Calculating probability often involves finding the ratio of successful outcomes to the total number of possible outcomes. With combinatorial problems, such as picking parts from a bin, using combinations helps us calculate how many successful outcomes exist. Let's break it down:
  • Calculate all possible selections: Use \(\binom{50}{10}\) to find total ways to sample parts.
  • Determine successful outcomes: Focus only on samples with 4 or 5 defective parts by calculating \(\binom{5}{4} \times \binom{45}{6}\) and \(\binom{5}{5} \times \binom{45}{5}\).
  • Add these outcomes: This gives us the total count of successful cases.
Remember, understanding these probabilities allows us to manage quality assurance and predict outcomes effectively in real-world situations.

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Most popular questions from this chapter

In control replication, cells are replicated over a period of two days. Not until mitosis is completed can freshly synthesized DNA be replicated again. Two control mechanisms have been identified-one positive and one negative. Suppose that a replication is observed in threc cells. Let \(A\) denote the event that all cells are identified as positive, and let \(B\) denote the event that all cells are negative. Describe the sample space graphically and display each of the following events: (a) A (b) \(B\) (c) \(A \cap B\) (d) \(A \cup B\)

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