Question

Components of a certain type are shipped to a supplier in batches of ten. Suppose that 47% of all such batches contain no defective components, 31% 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 54981 one defective component .290113 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 .44 1847 two defective components 558153

Answer 1

Answer:

Given that,

$\rightarrow$ Components of a certain type are shipped to a supplier in batches of ten. Suppose that 47% of all such batches contain no defective components, 31% contain one defective component, and 22% contain two defective components.

$\rightarrow$ 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:

$\rightarrow$ Let,

$B_{0}$ shows the event that batch with zero defective is selected.

$B_{1}$ shows the event that batch with one defective is selected.

$B_{2}$ shows the event that batch with two defective is selected.

$\rightarrow$ From the given information we have,

$P(B_{0})=0.47$

P(B1) = 0.31

$P(B_{2})=0.22$

(a).

Neither tested component is defective:

$\rightarrow$ Let D shows the event that no defective is found out of 2 items.

If a batch contains no defective so,

$P(D|B_{0})=1$

If the batch contains one defective (That is 9 non-defective) and our selected components have no defective then,

$P(D|B_{1})=\frac{\binom{9}{2}}{\binom{10}{2}}=0.8$

If the batch contains two defectives (That is 8 non-defective) and our selected components have no defective then,

$P(D|B_{2})=\frac{\binom{8}{2}}{\binom{10}{2}}=\frac{56}{90}=0.6222$

$\rightarrow$ Using the law of total probability, the probability that selected components have no defective is,

$P(D)=P(D|B_{0})P(B_{0})+P(D|B_{1})P(B_{1})+P(D|B_{2})P(B_{2})$

$=1\times 0.47+0.8\times 0.31+0.6222\times 0.22$

=0.47+0.248+0.1369

P(D)=0.8549

Then,

No defective components:

$P(B_{0}|D)=\frac{P(D|B_{0})P(B_{0})}{P(D)}$

$=\frac{1\times 0.47}{0.8549}$

=0.47/0.8549

=0.5497719

=0.5498 (Approximately)

Therefore, No defective components are 0.5498.

One defective component:

$P(B_{1}|D)=\frac{P(D|B_{1})P(B_{1})}{P(D)}$

$=\frac{0.8\times 0.31}{0.8549}$

=0.248/0.8549

=0.2900924

=0.2901 (Approximately)

Therefore, one defective component is 0.2901.

Two defective components:

$P(B_{2}|D)=\frac{P(D|B_{2})P(B_{2})}{P(D)}$

$=\frac{0.6222\times 0.22}{0.8549}$

=0.1369/0.8549

=0.16013567

=0.1601 (Approximately)

Therefore, two defective components are 0.1601.

(b).

One of the two tested components is defective:

$\rightarrow$ Let D shows the event that one defective is found out of 2 items.

If a batch contains no defective so,

$P(D|B_{0})=0$

If the batch contains one defective (That is 9 non-defective) and our selected components have one defective then,

$P(D|B_{1})=\frac{\binom{9}{1}\binom{1}{1}}{\binom{10}{2}}=0.2$

If the batch contains two defectives (That is 8 non-defective) and our selected components have one defective then,

$P(D|B_{2})=\frac{\binom{8}{1}\binom{2}{1}}{\binom{10}{2}}=\frac{16}{45}=0.3556$

$\rightarrow$ Using the law of total probability, the probability that selected components have no defective is,

$P(D)=P(D|B_{0})P(B_{0})+P(D|B_{1})P(B_{1})+P(D|B_{2})P(B_{2})$

$=0\times 0.47+0.2\times 0.31+0.3556\times 0.22$

=0+0.062+0.0782

P(D)=0.1402

Then,

No defective components:

$P(B_{0}|D)=\frac{P(D|B_{0})P(B_{0})}{P(D)}$

$=\frac{0\times 0.47}{0.1402}$

=0

Therefore, No defective components are 0.

One defective component:

$P(B_{1}|D)=\frac{P(D|B_{1})P(B_{1})}{P(D)}$

$=\frac{0.2\times 0.31}{0.1402}$

=0.062/0.1402

=0.44222539

=0.4422 (Approximately)

Therefore, one defective component is 0.4422.

Two defective components:

$P(B_{2}|D)=\frac{P(D|B_{2})P(B_{2})}{P(D)}$

$=\frac{0.3556\times 0.22}{0.1402}$

=0.0782/0.1402

=0.55777461

=0.5578 (Approximately)

Therefore, two defective components are 0.5578.