An integrated circuit factory has three machines X, Y, and Z. Test one integrated circuit produced by each machine. Either a circuit is acceptable (a) or it fails (f). An observation is a sequence of three test results corresponding to the circuits for machines X, Y, and Z, respectively. For example, aaf is the observation that the circuits from X and Y pass the test and the circuit from Z fails the test. a. What are the elements of the sample space of this experiment?
b. What are the elements of the sets:

ZF={circuit from Z fails},
XA ={circuit from X is acceptable}.

c. Are ZF and XA mutually exclusive?
d. Are ZF and XA collectively exhaustive?
e. What are the elements of the sets C={more than one circuit acceptable}, D={at least two circuits fail}.
f. Are C and D mutually exclusive?
g. Are C and D collectively exhaustive?

Kindly check explanation

Step-by-step explanation:

Acceptable = a

Fails = f

For machines X, Y, Z ;

Sample space, S = {aaa, aaf, aff, fff, faa, afa, ffa, faf}

Total elements in sample space = 8

B.)

Elements of the set:

ZF={circuit from Z fails},

XA ={circuit from X is acceptable}.

ZF = {aaf, aff, fff, faf}

XA = {aaa, aaf, aff, afa}

C.)

c. Are ZF and XA mutually exclusive?

Check if ZF n XA = ∅

ZF n XA = {aaf, aff}

Hence, ZF and XA are not mutually exclusive

d. Are ZF and XA collectively exhaustive?

To be collectively exhaustive :

ZF u XA = Sample space

ZF u XA = {aaa, aaf, aff, afa, faf, fff}

{aaa, aaf, aff, afa, faf, fff} ≠ sample space

Hence, ZF and ZA are not collectively exhaustive.

e. What are the elements of the sets

C={more than one circuit acceptable},

C = {aaa, aaf, faa, afa}

D={at least two circuits fail}.

D = {ffa, faf, aff, fff}

f. Are C and D mutually exclusive?

Check if C n D = ∅

C n D = ∅

Hence, A and D are mutually exclusive

g. Are C and D collectively exhaustive?

To be collectively exhaustive :

C u D = Sample space

C u D = {aaa, aaf, aff, afa, faf, fff}

{aaa, aaf, faa, ffa, aff, afa, faf, fff} = sample space

Hence, C and D are collectively exhaustive.

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.