Answer:
(a) The probability that an item has an inspection is 0.1984.
(b) The probability that an item is defective given that it has an inspection is 0.2581.
Step-by-step explanation:
The law of total probability states that, if B₁, B₂, B₃,... are part of a sample space S, then for any event A,
[tex]P(A)=\sum\limits_{i}{P(A|B_{i})\cdot P(B_{i})}[/tex]
The condition probability of an event A given that another event B has already occurred is:
[tex]P(A|B)=\frac{P(B|A)P(A)}{P(B)}[/tex]
Denote the events as follows:
X = an item is defective
Y = the item is being inspected by the inspector.
The information provided is:
[tex]P (X) = 0.08\\P (Y |X)=0.64\\P(Y|X^{c})=0.16[/tex]
(a)
Compute the probability that an item has an inspection as follows:
Use the law of total probability:
[tex]P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})[/tex]
[tex]=(0.64\times 0.08)+(0.16\times (1-0.08))\\=0.0512+0.1472\\=0.1984[/tex]
Thus, the probability that an item has an inspection is 0.1984.
(b)
Compute the probability that an item is defective given that it has an inspection as follows:
Use the condition probability:
[tex]P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)}[/tex]
[tex]=\frac{0.64\times 0.08}{0.1984}\\\\=0.2581[/tex]
Thus, the probability that an item is defective given that it has an inspection is 0.2581.