Question

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defective components, 28% contain one defective component, and 22% 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? (Round your answers to four decimal places.)

(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

Answer 1

a)

P(0 defective)=P(0 defective batch and 0 defective)+P(1 defective batch and 0 defective)+P(2 defective batch and 0 defective)

= $0.5*1+0.28*\binom{1}{0}\binom{9}{2}/\binom{10}{2}+0.22*\binom{2}{0}\binom{8}{2}/\binom{10}{2}$

=0.5*1+0.28*0.8+0.22*0.6222=0.8609

hence given 0 defective:

P(no defective component batch)==0.5*1/0.8609=0.5808

P(one defective component batch)==0.28*0.8/0.8609=0.2602

P(two defective component batch)==0.22*0.6222/0.8609=0.1590

b)

as above :

P(1 defective)=P(0 defective batch and 1 defective)+P(1 defective batch and 1 defective)+P(2 defective batch and 1 defective)

=

$0.5*0+0.28*\binom{1}{1}\binom{9}{1}/\binom{10}{2}+0.22*\binom{2}{1}\binom{8}{1}/\binom{10}{2}$

=0.5*0+0.28*0.2+0.22*0.3556=0.1342

hence given 1 defective

P(no defective component batch)==0.5*0/0.1342=0

P(one defective component batch)==0.28*0.2/0.1342=0.4172

P(two defective component batch)==0.22*0.3556/0.1342=0.5828