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Chapter 5: Problem 69

Suppose that you have just received a shipment of 20 modems. Although you don鈥檛 know this, 3 of the modems are defective. To determine whether you will accept the shipment, you randomly select 4 modems and test them. If all 4 modems work, you accept the shipment. Otherwise, the shipment is rejected. What is the probability of accepting the shipment?

Short Answer

Expert verified
The probability of accepting the shipment is 0.491.

Step by step solution

01

- Understand the Problem

We need to find the probability of accepting a shipment of 20 modems when selecting 4 modems at random and testing them. The shipment is accepted only if all 4 modems tested are non-defective.
02

- Total Number of Ways to Choose 4 Modems

First, calculate the total number of ways to choose 4 modems out of 20. This can be done using the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Where: - \(n\) is the total number of modems (20) - \(k\) is the number of modems to be chosen (4) \[ \binom{20}{4} = \frac{20!}{4!(20-4)!} = 4845 \]
03

- Number of Ways to Choose 4 Non-Defective Modems

Next, calculate the number of ways to choose 4 non-defective modems from the 17 non-defective modems. This is again done using the combination formula: \[ \binom{17}{4} = \frac{17!}{4!(17-4)!} = 2380 \]
04

- Calculate the Probability

To find the probability that all 4 modems chosen are non-defective, divide the number of ways to choose 4 non-defective modems by the total number of ways to choose 4 modems: \[ P(\text{All 4 modems work}) = \frac{\binom{17}{4}}{\binom{20}{4}} = \frac{2380}{4845} \ \therefore P(\text{Accept the shipment}) = 0.491 \]

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Key Concepts

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

combinatorial analysis
Combinatorial analysis involves the study of counting, arrangement, and combination of objects. It's a fundamental part of probability theory. Here, we analyze different ways to select and arrange objects according to specific rules.
In our example of modems, we use combinatorial analysis to count how many ways we can choose 4 modems out of 20. This helps us determine the total number of possible outcomes.
Overall, combinatorial analysis is crucial for solving problems involving selection and arrangement, making it easier to calculate probabilities accurately.
binomial coefficient
The binomial coefficient, denoted as \(\binom{n}{k}\), is a key element in calculating combinations. It tells us how many ways we can choose \(k\) items from \(n\) items without considering the order.
In our example with modems, we use the binomial coefficient to find out how many ways we can choose 4 modems from 20. The formula for binomial coefficient is: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
  • \(n!\) represents the factorial of n
  • \(k\) is the number of items chosen
  • The denominator \((n-k)!\) accounts for the items not chosen
Using the binomial coefficient helps simplify complex counting problems, enabling easier probability calculations.
sampling without replacement
Sampling without replacement means selecting items from a population where each item can only be chosen once. Once selected, the item is not returned back into the population.
In our modem example, we randomly select 4 modems from a shipment of 20 without putting them back. This type of sampling affects the probabilities because the number of available items decreases with each selection.
It's important to differentiate between sampling with and without replacement when calculating probabilities, as it greatly impacts the outcome. In our problem, since defective modems are not replaced, it accurately reflects real-world scenarios where items are not reused.

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Most popular questions from this chapter

You are dealt 5 cards from a standard 52-card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering the following questions: (a) How many ways can 5 cards be selected from a 52-card deck? (b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck? (c) The remaining 2 cards must be different from the 3 chosen and different from each other. For example, if we drew three kings, the 4th card cannot be a king. After selecting the three of a kind, there are 12 different ranks of card remaining in the deck that can be chosen. If we have three kings, then we can choose twos, threes, and so on. Of the 12 ranks remaining, we choose 2 of them and then select one of the 4 cards in each of the two chosen ranks. How many ways can we select the remaining 2 cards? (d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and two cards that are not like?

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