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Suppose you have just received a shipment of 20 modems. Although you don't 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
\( \frac{476}{969} \)

Step by step solution

01

- Understand the Problem

We need to find the probability of selecting 4 modems out of 20, without selecting any defective ones. There are 3 defective modems in the shipment.
02

- Total Number of Ways to Choose 4 Modems

Calculate the total number of ways to choose 4 modems out of 20. This is a combination problem. The formula for combinations is given by \[ C(n, k) = \frac{n!}{k!(n-k)!} \]Here, n = 20 and k = 4. So, we have \[ C(20, 4) = \frac{20!}{4!(20-4)!} = \frac{20!}{4! \cdot 16!} \]Using a calculator to find the value gives us \[ C(20, 4) = 4845 \]
03

- Number of Ways to Choose 4 Non-Defective Modems

Next, calculate the number of ways to choose 4 non-defective modems out of the 17 non-defective ones (since there are 3 defective and 20 total modems). Again, using the combination formula: \[ C(17, 4) = \frac{17!}{4!(17-4)!} = \frac{17!}{4! \cdot 13!} \]Using a calculator to find the value gives us \[ C(17, 4) = 2380 \]
04

- Calculate the Probability

The probability of all selected modems working (no defective ones) is calculated by dividing the number of favorable outcomes by the total number of outcomes: \[ P(\text{all 4 work}) = \frac{C(17, 4)}{C(20, 4)} = \frac{2380}{4845} \]Simplifying this fraction: \[ P(\text{all 4 work}) = \frac{476}{969} \]

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Key Concepts

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

Combinations in Probability
Combinations are a key concept in probability, especially when the order of selection doesn't matter. In our problem with modems, the task is to select 4 out of 20 modems.
The formula for combinations is: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] Here, \ {n\} is the total number of items (20 modems) and \ {k\} is the number of items to choose (4 modems).
Using this formula gives us the total number of ways to choose 4 modems from 20. This understanding is crucial because it forms the base of calculating probabilities in our scenario.
Defective Items
In probability exercises, defective items represent the unfavored outcomes we want to avoid. For our shipment of 20 modems, there are 3 defective ones.
When calculating probabilities, we need to know both the number of defective items and the total items. This helps us determine the favorable conditions. In our modem case, avoiding these 3 defective items is essential to determining if the shipment is accepted.
This concept is important because it directly influences the number of favorable outcomes available, impacting the probability calculations.
Calculating Probability
Calculating probability involves determining the ratio of favorable outcomes to the total number of possible outcomes.
For our modem problem, the probability of selecting 4 working modems is given by: \[ P(\text{all 4 work}) = \frac{C(17, 4)}{C(20, 4)} \] Here, \ {C(17, 4)\} represents the number of ways to choose 4 working modems from the 17 non-defective ones.
Likewise, \ {C(20, 4) \} is the total number of ways to choose any 4 modems from the 20 available.
Dividing these values simplifies the complex task of finding probability down to a manageable calculation. This method ensures accuracy and clarity in our solution.
Non-Defective Items
Non-defective items are those in the group that meet the quality standards. In this exercise, there are 17 non-defective modems out of the total 20.
Calculating the favorable outcomes requires us to focus on the non-defective items. For selecting 4 working modems, we use the combinations formula to find the number of ways to pick from these 17 good modems: \[ C(17, 4) = \frac{17!}{4!(17-4)!} \]
This gives us the total possible ways to pick 4 non-defective ones. This number is critical for comparing against the total combinations, thereby forming the basis of our probability calculations.
Focusing on non-defective items ensures that only the acceptable configurations are counted in our favorable outcomes.

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Most popular questions from this chapter

Due to a manufacturing error, three cans of regular soda were accidentally filled with diet soda and placed into a 12-pack. Suppose that two cans are randomly selected from the case. (a) Determine the probability that both contain diet soda. (b) Determine the probability that both contain regular soda. Would this be unusual? (c) Determine the probability that exactly one is diet and one is regular.

Suppose a computer chip company has just shipped 10,000 computer chips to a computer company. Unfortunately, 50 of the chips are defective. (a) Compute the probability that two randomly selected chips are defective using conditional probability. (b) There are 50 defective chips out of 10,000 shipped. The probability that the first chip randomly selected is defective is \(\frac{50}{10,000}=0.005=0.5 \% .\) Compute the probability that two randomly selected chips are defective under the assumption of independent events. Compare your results to part (a). Conclude that, when small samples are taken from large populations without replacement, the assumption of independence does not significantly affect the probability.

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A committee consists of four women and three men. The committee will randomly select two people to attend a conference in Hawaii. Find the probability that both are women.

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