Question

Units of a manufactured product that are to be inspected are known to be subject to three types of defects. Six percent of the units have surface finish flaws, 4% have improperly fixed ground wires, and 3% have base mounts that are not level. It is known that .25% of the units have all three defect types and .50% have surface finish flaws and improperly fixed ground wires. In addition, we know that .50% have the combination of only surface finish flaws and base mounts that are not level, and 1.5% have only the unlevel base mount defect.

a. What is the probability that a unit selected at random will have both an improperly fixed ground wire and a base mount that is not level?

b. What is the probability that a unit selected at random will have only the improperly fixed ground wire defect?

c. What is the probability that a unit selected at random will have only the surface flaw defect?

d. What is the probability that a unit selected at random will be defect free?

I am having trouble, especially because it says that 3% have the base mount defect then later says 1.5% have the base mount defect. Please guide me through your thinking, thank you :)

Answer 1

here let probability of having surface finish flaws , improperly fixed ground wires and base mounts that are not level are A,B and C

P(A)=			0.06
P(B)=			0.04
P(C)=			0.03
P(AnB)=			0.005

P(AnC)=P(A)+P(C)-P(A U C)=

0.005

P(AnBnC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AuBuC)=

0.0025

here as P(only unlevel base mount defect)= P(C n A^c n B^c)=P(C)-P(A n C)-P(B n C)+P(A n B n C)

=0.03-0.005-P(B n C)+0.0025 =0.015

P(B n C) =0.0125

a)

probability that a unit selected at random will have both an improperly fixed ground wire and a base mount that is not level

=P( B n C) =0.0125

b)

probability that a unit selected at random will have only the improperly fixed ground wire defect

=P(A^cn B n C^c)=P(B)-P(A n B)-P(B n C) +P(A n B nC )=0.025

c)

probability that a unit selected at random will have only the surface flaw defect =P(A n B^c n C^c)

=P(A)-P(A n B)-P(A n C)+P(A n B n C)=0.06-0.005-0.005+0.0025 =0.0525

d)

at least one =P(AUBUC)=

P(A)+P(B)+P(C )-P(AnB)-P(BnC)-P(AnC)+P(AnBnC)=

0.11

hence P(no defect) =1-0.11 =0.89