Problem 4E

Three times each day, a quality engineer samples a component from a recently manufactured batch and tests it. Each part is classified as conforming (suitable for its intended use), downgraded (unsuitable for the intended purpose but usable for another purpose), or scrap (not usable). An experiment consists of recording the categories of the three parts tested in a particular day.

a. List the 27 outcomes in the sample space.

b. Let A be the event that all the parts fall into the same category. List the outcomes in A.

c. Let B be the event that there is one part in each category. List the outcomes in B.

d. Let C be the event that at least two parts are conforming. List the outcomes in C.

e. List the outcomes in A ∩ C.

f. List the outcomes in A ∪ B.

g. List the outcomes in A ∩ Cc.

h. List the outcomes in Ac ∩ C.

i. Are events A and C mutually exclusive? Explain.

j. Are events B and C mutually exclusive? Explain.

Solution 4E

Step1 of 6:

We have a quality engineer samples a component from a recently manufactured batch and tests it 3 times each day. Each part is classified as

1).Conforming

2).Downgraded and

3).Scrap.

An experiment consists of recording the categories of the three parts tested in a particular day.

We need to find,

a).We need to List the 27 outcomes in the sample space.

b).Let A be the event that all the parts fall into the same category. We need to List the outcomes in A.

c). Let B be the event that there is one part in each category. We need to List the outcomes in B.

d).Let C be the event that at least two parts are conforming. We need to List the outcomes in C.

e).We need to List the outcomes in A ∩ C.

f).We need to List the outcomes in A ∪ B.

g).We need to List the outcomes in A ∩ Cc.

h).We need to List the outcomes in Ac ∩ C.

i).We need to check Are events A and C mutually exclusive? Explain.

j).We need to check Are events B and C mutually exclusive? Explain.

Step2 of 6:

a).

Here, we have

1).Conforming(C)

2).Downgraded(D) and

3).Scrap(S).

The possible number of 27 outcomes in the sample space is S = {CCC, CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS, DCC, DCD, DCS, DDC, DDD, DDS, DSC, DSD, DSS, SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}.

b).

Here we have assumed that A be the event that all the parts fall into the same category. That is

1).Conforming(C)

2).Downgraded(D) and

3).Scrap(S).

These all parts fall into the same part A and the are given by

A = {CCC, DDD, SSS}

Step3 of 6:

c).

Here we have assumed that B be the event that there is one part in each category. That is

1).Conforming(C)

2).Downgraded(D) and

3).Scrap(S).

In this three parts one part must be there in each category and it is given by

B = {CDS,CSD, DCS, DSC, SCD, SDC}.

d).

Here we have assumed that C be the event that at least two parts are conforming. That is

1).Conforming(C)

2).Downgraded(D) and

3).Scrap(S).

In this three parts there must be 2 C’s in each category and it is given by

C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC}.

Step4 of 6:

e).

We have A = {CCC, DDD, SSS} and C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC}.

Now,

A ∩ C = {common elements in group A and C}

There are only one element is common in group A and B that is CCC

= {CCC}

Therefore, A ∩ C = {CCC}.

f).

We have A = {CCC, DDD, SSS} and B = {CDS,CSD, DCS, DSC, SCD, SDC}.

Now,

A ∪ B = {all the elements in group A and group B}

= {CCC, DDD, SSS, CDS,CSD, DCS, DSC, SCD, SDC}.

Therefore, A ∪ B = {CCC, DDD, SSS, CDS,CSD, DCS, DSC, SCD, SDC}.

Step5 of 6:

g).

We have S = S = {CCC, CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS, DCC, DCD, DCS, DDC, DDD, DDS, DSC, DSD, DSS, SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}.

A = {CCC, DDD, SSS} and C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC}.

Now,

Cc = {write elements in sample space S that the outcomes that are not in C}

Cc = {CDD, CDS, CSD, CSS, DCD, DCS, DDC, DDD, DDS, DSC, DSD,

DSS, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}

Hence,

A ∩ Cc = {common elements in group A and Cc}

= {DDD, SSS}

Therefore, A ∩ Cc = {DDD, SSS}.

h).

We have S = {CCC, CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS, DCC, DCD, DCS, DDC, DDD, DDS, DSC, DSD, DSS, SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD, SSS}.

A = {CCC, DDD, SSS} and C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC}.

Now,

Ac = {write elements in sample space S that the outcomes that are not in A}

Ac = {CCD, CCS, CDC, CDD, CDS, CSC, CSD, CSS, DCC, DCD, DCS, DDC,

DDS, DSC, DSD, DSS, SCC, SCD, SCS, SDC, SDD, SDS, SSC, SSD,}

Hence,

Ac ∩ C = {common elements in group A and Cc}

= {CCD,CCS, CDC, CSC, DCC, SCC}

Therefore, Ac ∩ C = {CCD,CCS, CDC, CSC, DCC, SCC}.

Step6 of 6:

i).

We have A = {CCC, DDD, SSS} and C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC}.

Now,

A ∩ C = {CCC}. [therefore, From part(e)]

Hence, A ∩ C 0.

Therefore, the events A and C are not mutually Exclusive events.

j).

We have B = {CDS,CSD, DCS, DSC, SCD, SDC} and C = {CCD, CCS, CDC, CSC, DCC, SCC, CCC}.

Now,

B ∩ C = {common elements in group B and C}

= {0}

Hence, B ∩ C = 0.

Therefore, the events B and C are mutually Exclusive events.