In: Statistics and Probability
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 :)
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 Ac n Bc)=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(Acn B n Cc)=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 Bc n Cc)
=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