Question

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 79% of all such batches contain no defective components, 15% contain one defective component, and 6% 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.

0:

1:

2:
(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.]

0:

1:

2:

NEED this ASAP

Thanks.

Answer 1

a)

a	b	a*b	ab/Σab
P(x)	P(no defect|x)	P(x and no defect)	P(x|no defect)
0.79	1.0000	0.7900	0.8339
0.15	0.8000	0.1200	0.1267
0.06	0.6222	0.0373	0.0394
	Σab =	0.9473

P(no defective batch given no defective)=P(no defective batch and no defective)/P(no defective)=0.8339

P(one defective batch given no defective)=P(one defective batch and no defective)/P(no defective)=0.1267

P(two defective batch given no defective)=P(two defective batch and no defective)/P(no defective)=0.0394

b)

a	b	a*b	ab/Σab
P(x)	P(one defect|x)	P(x and one defect)	P(x|one defect)
0.79	0.0000	0.0000	0.0000
0.15	0.2000	0.0300	0.5844
0.06	0.3556	0.0213	0.4156
	Σab =	0.0513

P(no defective batch given one defective)=P(no defective batch and one defective)/P(one defective)=0

P(one defective batch given one defective)=P(one defective batch and one defective)/P(one defective)=0.5844

P(two defective batch given one defective)=P(two defective batch and one defective)/P(one defective)=0.4156