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Chapter 4: Problem 105

A shipping container contains seven complex electronic systems. Unknown to the purchaser, three are defective. Two of the seven are selected for thorough testing and are then classified as defective or nondefective. What is the probability that no defectives are found?

Short Answer

Expert verified
Answer: The probability of selecting two non-defective electronic systems is 2/7 or approximately 0.286.

Step by step solution

01

Calculate the total number of combinations

First, we need to find the total number of ways to select two electronic systems out of seven. This can be done using the combination formula: \[C(n, k) = \frac{n!}{k!(n-k)!}\] In this case, we have \(n = 7\) and \(k = 2\), so the total number of combinations is: \[C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7\times6}{2\times1} = 21\]
02

Calculate the number of favorable combinations

Now, we need to find the number of ways to select two non-defective systems out of the four non-defective ones available. Using the same combination formula \[C(n, k) = \frac{n!}{k!(n-k)!}\] This time, we have \(n = 4\) and \(k = 2\), so the number of favorable combinations is: \[C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4\times3}{2\times1} = 6\]
03

Calculate the probability

To find the probability of selecting two non-defective systems, we need to divide the number of favorable combinations by the total number of combinations. So, we have: \[P(\text{no defectives}) = \frac{\text{favorable combinations}}{\text{total combinations}} = \frac{6}{21} = \frac{2}{7}\] Thus, the probability of selecting two non-defective electronic systems is \(\frac{2}{7}\) or approximately 0.286.

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