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A group of four components is known to contain two

Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer ISBN: 9780495110811 47

Solution for problem 3E Chapter 3

Mathematical Statistics with Applications | 7th Edition

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Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

Mathematical Statistics with Applications | 7th Edition

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Problem 3E

Problem 3E

A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two defectives are located. Once she locates the two defectives, she stops testing, but the second defective is tested to ensure accuracy. Let Y denote the number of the test on which the second defective is found. Find the probability distribution for Y .

Step-by-Step Solution:
Step 1 of 3

Solution 3E

Step1 of 2:

Let us consider an event ‘D’ be the defective components.

Let event ‘G’ no defective components.

We have 2 defectives out of 4 components.

We need to find the probability distribution for Y.


Step2 of 2:

Let Y denote the number of the test on which the second defective is found and it is given as:

P(both defective are found in the first two inspections) = P(Y = 2)

 

                                                                                          = P(DD)

                                       

                                                                                          = P(Prob. Of first defective)P(Prob.Of

                                                                                                                               second defective)

                                                                                          =

                                                                               =

Hence, P(both defective are found in the first two inspections) = .

Similarly,

For Y = 3, we need to break it into two cases that is GDD and DGD.

P(Y = 3) =  2C1

             = (2)

                                                   =

Hence, P(Y = 3) = .

Similarly,

For Y = 3, we need to break it into three cases that is GGDD, DGGD, GDGD.

 

P(Y = 4) =  3C1

              = (3)

                                                   =

Hence, P(Y = 4) = .


Step 2 of 3

Chapter 3, Problem 3E is Solved
Step 3 of 3

Textbook: Mathematical Statistics with Applications
Edition: 7
Author: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer
ISBN: 9780495110811

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A group of four components is known to contain two