A manufacturing company changes an acceptance scheme on items from a production line before they are shipped. An inspector takes 1 item at random from a box of21 items, inspects it, and then replaces it in the box; a second inspector does likewise. Finally, a third inspector goes through the same procedure. If any of the threeinspectors find a defective, the entire box is sent back for 100% screening. If no defectives are found, the box is shipped.(a) What is the probability that a box containing 4 defectives will be shipped?(b) What is the probability that a box containing only 1 defective will be sent back for screening?

Answered: A manufacturing company changes an…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A manufacturing company changes an acceptance scheme on items from a production line before they are shipped. An inspector takes 1 item at random from a box of
21 items, inspects it, and then replaces it in the box; a second inspector does likewise. Finally, a third inspector goes through the same procedure. If any of the three
inspectors find a defective, the entire box is sent back for 100% screening. If no defectives are found, the box is shipped.
(a) What is the probability that a box containing 4 defectives will be shipped?
(b) What is the probability that a box containing only 1 defective will be sent back for screening?

Expert Solution

Step 1

To find the probability of box containing 4 defectives will be shipped. Let random variable X represents the number of defective item among the tested one item. Therefore, the box will be shipped there is no defective. So, we know x=0.

Therefor, we have to find the probability of the hypergeometric random variable X is given by,

$\begin{array}{rcl} h (x; N, n, k) & = & \frac{(\begin{matrix} k \\ x \end{matrix}) (\begin{matrix} N - k \\ n - x \end{matrix})}{(\begin{matrix} N \\ n \end{matrix})} \end{array}$

We have, k=4, n=1, x=0 and N=21

Therefore, probability that box containing 4 defective will be shipped is,

$\begin{array}{rcl} h (x; N, n, k) & = & \frac{(\begin{matrix} k \\ x \end{matrix}) (\begin{matrix} N - k \\ n - x \end{matrix})}{(\begin{matrix} N \\ n \end{matrix})} \\ = & \frac{(\begin{matrix} 4 \\ 0 \end{matrix}) (\begin{matrix} 21 - 4 \\ 1 - 0 \end{matrix})}{(\begin{matrix} 21 \\ 1 \end{matrix})} \\ = & \frac{\frac{4!}{0! (4 - 0)!} \cdot \frac{17!}{1! (17 - 1)!}}{\frac{21!}{1! (21 - 1)!}} \\ = & \frac{1 \times 17}{21} \\ = & 0.81 \end{array}$

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