Components of a certain type are shipped to a supplier in batches of ten. Suppose that 52% of all such batches contain no defective components, 28% contain one defective component, and 20% contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?
(a) Neither tested component is defective.
no defective components :
one defective component :
two defective components :
(b) One of the two tested components is defective. [Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]
no defective components :
one defective component :
two defective components :

P ( B0 / D0 ) = 0.59877

P ( B1 / D0 ) = 0.25793

P ( B2 / D0 ) = 0.14329

Step-by-step explanation:

Given:

-  0 be the event that the batch has 0 defectives = (0 ) = 0.52

- 1 be the event that the batch has 1 defectives = (1 ) = 0.28

- 2 be the event that the batch has 2 defectives = (2 ) = 0.2

- Two components are selected

Find:

What are the probabilities associated with 0, 1, and 2 defective components being in the batch under each of the following conditions?

(a) Neither tested component is defective.

Solution:

Let 0 be the event that neither selected component is defective.

- The event 0 can happen in three different ways:

(i) Our batch of 10 is perfect, and we get no defectives in  our sample of two;

                   P(i) = P(B0) = 0.52

(ii) Our batch of 10 has 1 defective, but our sample of two misses them;

                  P ( no defect / B1 ) = P ( no defect ) / P ( B 1 )

                                                  = 9C2 / 10C2 = 0.8

                  P ( ii ) = 0.28*0.8 = 0.224

(iii) Our batch  has 2 defective, but our sample misses them.

                 P ( no defect / B2 ) = P ( no defect ) / P ( B 2 )

                                                  = 8C2 / 10C2 = 56/90

                  P ( iii ) = 0.2*56/90 = 0.124444

- Then,

                 P(Do) = P(i) + P(ii) + P(iii)

                 P(Do) = 0.52 + 0.224 + 0.124444 = 977/1125

We use the general conditional probability formula:

                P ( B0 / D0 ) = P ( B0 & D0 ) / P( D0 )

                P ( B0 / D0 ) = 0.52*1125 / 977 = 0.59877

                P ( B1 / D0 ) = P ( B1 & D0 ) / P( D0 )

                P ( B1 / D0 ) = 0.224*1125 / 977 = 0.25793

                P ( B2 / D0 ) = P ( B2 & D0 ) / P( D0 )

                P ( B2 / D0 ) = 0.12444*1125 / 977 = 0.14329