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Chapter 5: Problem 92

Improperly wired control panels were mistakenly installed on two of eight large automated machine tools. It is uncertain which of the machine tools have the defective panels, and a sample of four tools is randomly chosen for inspection. What is the probability that the sample will include no defective panels? Both defective panels?

Short Answer

Expert verified
Answer: The probability of selecting a sample with no defective panels is \(\dfrac{3}{14}\), and the probability of selecting a sample with both defective panels is also \(\dfrac{3}{14}\).

Step by step solution

01

Understand the given problem

We have 8 machine tools, with 2 defective and 6 non-defective ones. We need to find the probability of selecting a sample of 4 tools with no defective panels and both defective panels.
02

Step 2:Use combinations to find selection options

The total number of ways to choose 4 tools out of 8 can be calculated using combinations, denoted as C(n, r), where n is the total number of elements and r is the number of elements we want to select. In our case, we want to choose 4 tools (r=4) out of 8 (n=8): Total options: C(8, 4) = \(\dfrac{8!}{4! (8-4)!} = 70\)
03

Calculate the probability of selecting no defective panels

First, we will calculate the probability of selecting a sample with no defective panels. To do this, we need to find the number of combinations that do not include any of the 2 defective panels. Since there are 6 non-defective panels and we need to choose 4, we have: No defective panels options: C(6, 4) = \(\dfrac{6!}{4! (6-4)!} = 15\) Now, to find the probability, we will divide the total possible selections by the total options: Probability (no defective panels) = \(\dfrac{15}{70} = \dfrac{3}{14}\)
04

Calculate the probability of selecting both defective panels

Next, we will calculate the probability of selecting a sample with both defective panels. To do this, we need to find the number of combinations that include the 2 defective panels and 2 non-defective panels. Since there are 6 non-defective panels and we need to choose 2, we have: Non-defective panels options: C(6, 2) = \(\dfrac{6!}{2! (6-2)!} = 15\) Then, we need to multiply this by 1 since both defective panels must be selected: Defective panels options: C(2, 2) = \(\dfrac{2!}{2! (2-2)!} = 1\) So, the total number of options for selecting both defective panels from the 4 inspected tools is 1*15 = 15. Now, to find the probability, we will divide the total possible selections with both defective panels by the total options: Probability (both defective panels) = \(\dfrac{15}{70} = \dfrac{3}{14}\) So, the probability of the sample including no defective panels is \(\dfrac{3}{14}\), and the probability of the sample including both defective panels is also \(\dfrac{3}{14}\).

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